Problem. Find all $f:\mathbb R_{>0}\to\mathbb R_{>0}$ with $\sqrt{\frac{x^2+f(y)^2}{2}}\ \ge\ \frac{f(x)+y}{2}\ \ge\ \sqrt{x f(y)}\qquad \forall x,y>0.$

Guess: $f(x)=x$ . Check: QM $\sqrt{(x^2+y^2)/2}\ge (x+y)/2 \ge \sqrt{xy}$ . Yes, works (QM-AM-GM).

Attack the right inequality (AM–GM part): $\frac{f(x)+y}{2}\ge \sqrt{x f(y)}$ .

Set $y = f(x)$ ? Hmm. Let’s think: RHS inequality with $x=y$ : $\frac{f(x)+x}{2}\ge \sqrt{x f(x)}$ — always true, no info.

Try: fix $x$ , let $y$ vary. $f(x) + y \ge 2\sqrt{x f(y)}$ , i.e. $f(y) \le \frac{(f(x)+y)^2}{4x}$ .

Left inequality: $\frac{x^2 + f(y)^2}{2} \ge \frac{(f(x)+y)^2}{4}$ , i.e. $2x^2 + 2f(y)^2 \ge (f(x)+y)^2$ .

Combine: from right ineq, $(f(x)+y)^2 \ge 4x f(y)$ . Left: $2x^2+2f(y)^2 \ge (f(x)+y)^2 \ge 4xf(y)$ , i.e. $2(x - f(y))^2 \ge 0$ . No info directly. Need to use both more cleverly.

Key idea: In the right inequality substitute swap roles. Let’s denote $a = f(x)$ , and consider setting $y$ such that… Let’s try $y \to$ specific.

Right inequality: $f(x) + y \ge 2\sqrt{x f(y)}$ for all $x,y$ .

Take $x$ with… Let’s try to show $f(y)\ge y$ and $f(y)\le y$ .

Step 1: Show $f(x)\ge x$ for all $x$ ? Suppose apply right inequality with $y$ replaced… Set $x=y$ : nothing.

Try: In right inequality set $x = f(y)$ : $\frac{f(f(y)) + y}{2}\ge \sqrt{f(y) f(y)} = f(y)$ . So $f(f(y)) + y \ge 2f(y)$ .

In left inequality set $x=f(y)$ : $\frac{f(y)^2 + f(y)^2}{2} \ge \left(\frac{f(f(y))+y}{2}\right)^2$ so $f(y) \ge \frac{f(f(y))+y}{2}$ , i.e. $2f(y)\ge f(f(y)) + y$ .

Combined with above: $f(f(y)) + y = 2f(y)$ for all $y$ . (*)

Nice! So equality holds in both when $x=f(y)$ .

So $f(f(y)) = 2f(y) - y$ for all $y>0$ . Since $f>0$ , we get $2f(y) - y > 0$ , so $f(y) > y/2$ .

Step 2: Iterate. Define $y_0=y$ , $y_{n+1} = f(y_n)$ . Then $y_{n+2} = 2y_{n+1} - y_n$ , i.e. arithmetic! Indeed $y_{n+2}-y_{n+1} = y_{n+1}-y_n$ . So with $d = f(y)-y$ , $y_n = y + nd$ , all must be positive for all $n\ge 0$ . So $y + nd>0$ for all $n$ , forcing $d\ge 0$ . So $f(y)\ge y$ for all $y$ .

Now need $f(y)\le y$ . Use inequalities again with $f(y)\ge y$ .

Right inequality: $f(x)+y \ge 2\sqrt{x f(y)} \ge 2\sqrt{xy}$ . Hmm that’s weaker.

Left inequality: $2x^2 + 2f(y)^2 \ge (f(x)+y)^2$ . With $f(x)\ge x$ : $2x^2+2f(y)^2 \ge (x+y)^2$ . Not immediate contradiction.

Let’s use the arithmetic-progression structure more. From $f(y_n) = y_{n+1} = y + (n+1)d$ . Also apply the left inequality with general $x$ and $y$ replaced by $y_n$ : hmm.

Better: use left inequality: for all $x,y$ : $\sqrt{\frac{x^2+f(y)^2}{2}} \ge \frac{f(x)+y}{2}$ . Since $f(x)\ge x$ , we get $\sqrt{\frac{x^2+f(y)^2}{2}}\ge \frac{x+y}{2}$ .

Take $y$ large relative… Let $f(y) = y + d(y)$ with $d(y)\ge 0$ . Need show $d\equiv 0$ .

Use right inequality: $f(x) + y \ge 2\sqrt{x f(y)}$ . Now use $f(x) = x + d(x)$ : $x + d(x) + y \ge 2\sqrt{x(y+d(y))}$ .

Hmm, we want an upper bound on $f$ . Use left inequality to bound $f(x)$ above: $2x^2 + 2f(y)^2 \ge (f(x)+y)^2$ , so $f(x) \le \sqrt{2x^2+2f(y)^2} - y$ . Minimize RHS over $y$ . Let $g(y)=\sqrt{2x^2+2f(y)^2}-y$ . If we can find $y$ where $f(y)$ small relative to $y$ … Since $f(y)\ge y$ : $g(y)\ge \sqrt{2x^2+2y^2}-y$ , minimized… Let’s just plug $f(y)=y$ hypothetically: $\sqrt{2x^2+2y^2}-y$ , minimize over $y$ : derivative $\frac{2y}{\sqrt{2x^2+2y^2}} - 1 = 0 \Rightarrow 4y^2 = 2x^2+2y^2 \Rightarrow y^2 = x^2$ , $y=x$ : value $\sqrt{4x^2}-x = x$ . So bound $f(x)\le x$ if $f(y)=y$ at $y=x$ … circular. Hmm.

Use the derived relation with $x=f(y)$ giving equality — maybe extract more from equality cases: equality in left inequality at $x=f(y)$ means… left inequality is $\sqrt{\frac{x^2+f(y)^2}{2}}\ge \frac{f(x)+y}{2}$ ; at $x=f(y)$ LHS $=f(y)$ exactly, and equality means $f(f(y))+y = 2f(y)$ — that’s the relation, fine.

Alternative: use both inequalities with $x$ and swapped variables to compare.

Let’s use: for all $x,y$ : (R) $f(x) + y \ge 2\sqrt{x f(y)}$ (L) $2x^2 + 2f(y)^2 \ge (f(x)+y)^2$ .

From (L): $f(x) \le \sqrt{2x^2+2f(y)^2} - y$ for all $y$ . From (R): $f(x)\ge 2\sqrt{xf(y)} - y$ .

So for all $x,y,z$ : $2\sqrt{x f(z)} - z \le \sqrt{2x^2 + 2f(y)^2} - y$ .

Now set $z = y$ : $2\sqrt{xf(y)} - y \le \sqrt{2x^2+2f(y)^2}-y$ → $4xf(y)\le 2x^2+2f(y)^2$ , trivial.

Set $x$ appropriately. Let $u=f(y)$ , $v = f(z)$ . Condition: $2\sqrt{xv} + y - z \le \sqrt{2x^2+2u^2}$ for all $x>0$ and all $y,z$ .

Choose $x = u$ (i.e. $x=f(y)$ ): LHS $= 2\sqrt{f(y)f(z)} + y - z$ , RHS $=2f(y)$ . So $2\sqrt{f(y)f(z)} \le 2f(y) + z - y \quad \forall y,z. \tag{A}$ With $f(y)=y+d(y)$ , $f(z)=z+d(z)$ .

Let’s test: suppose $d(y_0)>0$ for some $y_0$ . Use (A) with $y=y_0$ ? Then $2\sqrt{f(y_0)f(z)}\le 2f(y_0)+z-y_0$ . Set $t=f(y_0) = y_0+d_0$ . RHS $= 2y_0+2d_0+z-y_0 = y_0+2d_0+z$ . LHS $\ge 2\sqrt{t z}$ since $f(z)\ge z$ : $2\sqrt{(y_0+d_0)z}\le y_0+2d_0+z$ . Hmm by AM-GM $2\sqrt{(y_0+d_0)z} \le (y_0+d_0)+z$ … consistent. Need choose $z$ near $y_0+d_0$ : $z = y_0+d_0$ : LHS $\ge 2(y_0+d_0)$ , RHS $=y_0+2d_0+y_0+d_0 = 2y_0+3d_0$ . So $2y_0+2d_0 \le 2y_0+3d_0$ ✓. Not contradiction.

Also use (A) with roles: swap $y,z$ : $2\sqrt{f(z)f(y)}\le 2f(z)+y-z$ . Add to (A): $4\sqrt{f(y)f(z)}\le 2f(y)+2f(z)$ , trivial again. Hmm.

Add: $2\sqrt{f(y)f(z)} \le 2f(y)+z-y$ and $2\sqrt{f(y)f(z)}\le 2f(z)+y-z$ . So $4\sqrt{f(y)f(z)} \le 2f(y)+2f(z) \Rightarrow \text{trivial}.$ But individually: from (A), $2\sqrt{f(y)f(z)} - 2f(y) \le z-y$ . Write $f(y)=y+a$ , $f(z)=z+b$ with $a,b\ge 0$ . (A): $2\sqrt{(y+a)(z+b)} \le 2y+2a+z-y = y+2a+z$ . Hmm i.e. $2\sqrt{(y+a)(z+b)}\le (y+a)+(z+b) + (a - b)$ . By AM-GM the LHS $\le (y+a)+(z+b)$ with equality iff $y+a=z+b$ . So (A) is a real constraint when $a<b$ : need $2\sqrt{PQ} \le P+Q-(b-a)$ where $P=y+a,Q=z+b$ . I.e. $(\sqrt P-\sqrt Q)^2 \ge b-a$ .

So: for all $y,z$ : $(\sqrt{f(y)}-\sqrt{f(z)})^2 \ge d(z)-d(y)$ where $d(t)=f(t)-t$ .

Hmm, choose $z$ such that $f(z)=f(y)$ ? Then need $d(z)\le d(y)$ and vice versa… Not obviously available.

Better: choose $z$ with $f(z)$ close to $f(y)$ but $d(z) > d(y)$ . Use the AP structure: $f(y_n)=y+(n+1)d$ where $d=d(y)$ . Hmm, along the orbit $d(y_n) = d$ constant. So orbits have constant $d$ .

Alternative approach: maybe get $f$ nondecreasing / continuity? Let’s derive more constraints.

Let’s get a stronger inequality by choosing $x$ optimally in $2\sqrt{xv} + y - z \le \sqrt{2x^2+2u^2}$ .

Set $s = z - y$ . Inequality: $2\sqrt{x v} - s \le \sqrt{2x^2+2u^2}$ for all $x>0$ , where $u=f(y), v=f(z)$ .

We want to find the max over $x$ of $2\sqrt{xv} - \sqrt{2x^2+2u^2}$ and require $\le s$ .

Let me instead consider a cleaner path: use (R) with $y$ and (L) with the same pair to derive $f(x)$ pinned.

Actually maybe simpler: from (R), $f(x)\ge 2\sqrt{x f(y)} - y$ for all $y$ . Maximize RHS over $y$ using $f(y)\ge y$ : $\ge 2\sqrt{xy}-y$ , maximized at $y=x$ giving $2x - x = x$ . So $f(x)\ge x$ — consistent, recovers step 1 (alternative proof!). Indeed with $y=x$ : $f(x)\ge 2\sqrt{xf(x)}-x$ → $(\sqrt{f(x)}-\sqrt x)^2\ge ...$ let’s check: $f(x)+x\ge 2\sqrt{xf(x)}$ trivially true. Hmm no, so that’s trivial. Use $f(y)\ge y$ from Step 1 with $y$ chosen: $f(x)\ge 2\sqrt{xf(y)}-y$ . To get upper bound on $f$ we need (L).

From (L) with $y$ : $f(x) \le \sqrt{2x^2+2f(y)^2}-y$ . To prove $f(x)\le x$ we need some $y$ with $\sqrt{2x^2+2f(y)^2}-y \le x$ , i.e. $2x^2+2f(y)^2 \le (x+y)^2 = x^2+2xy+y^2$ , i.e. $2f(y)^2 \le -x^2+2xy+y^2$ . With $f(y)\ge y$ : need $2y^2 \le 2f(y)^2\le -x^2+2xy+y^2$ i.e. $y^2 - 2xy + x^2\le 0$ i.e. $(y-x)^2\le 0$ , so $y=x$ and $f(y)=y$ . Circular again — the inequalities are tight, need finer argument.

So both inequalities are tight; equality analysis needed. Let’s use the relation $f(f(y))=2f(y)-y$ more, i.e., $f$ acts on orbit as AP with common difference $d(y)$ . Note $f(y_n) = y_{n+1}$ and $d(y_n)=d(y)$ for all $n$ .

Also note from Step 1: $d(y)\ge0$ .

Now suppose $d(y_0) = d > 0$ for some $y_0$ . Then the orbit $y_0, y_0+d, y_0+2d,\dots$ all have $d(\cdot)=d$ , i.e. $f(y_0+nd) = y_0+(n+1)d$ .

Now plug into (L) or (R) with $x$ from the orbit and $y$ from the orbit, large $n$ ?

(L): $2x^2+2f(y)^2\ge (f(x)+y)^2$ . Take $x = y_0+md$ , $y=y_0+nd$ . Then $f(x)=y_0+(m+1)d$ , $f(y)=y_0+(n+1)d$ . Let $A=y_0$ , so: $2(A+md)^2 + 2(A+(n+1)d)^2 \ge (A+(m+1)d + A+nd)^2 = (2A+(m+n+1)d)^2$ .

Let $p = A+md$ , $q=A+nd$ . LHS $=2p^2+2(q+d)^2$ . RHS $=(p+d+q)^2$ . So need $2p^2+2(q+d)^2 \ge (p+q+d)^2$ . Set $Q=q+d$ : $2p^2+2Q^2 \ge (p+Q)^2$ ⟺ $(p-Q)^2\ge0$ ✓. Always fine. Hmm.

(R): $f(x)+y\ge 2\sqrt{xf(y)}$ : $(p+d)+q \ge 2\sqrt{p(q+d)}$ : same as $(p+Q)^2\ge 4pQ$ ✓. So orbit consistent — need to use non-orbit points, i.e., mixing different $y$ ’s with different $d$ .

So constraint (A): $(\sqrt{f(y)}-\sqrt{f(z)})^2 \ge d(z)-d(y)$ .

Hmm, more generally derive the full consequence: for all $x,y,z>0$ : $2\sqrt{x f(z)} - z \le f(x) \le \sqrt{2x^2+2f(y)^2}-y.$

Take $x$ in the orbit of $y_0$ : $x=y_0+md$ where $f(x) = x+d$ . Then upper bound: $x+d \le \sqrt{2x^2+2f(y)^2}-y$ for all $y$ . So $2x^2+2f(y)^2 \ge (x+d+y)^2$ .

Now pick $y$ with $d(y)$ small (we want to contradict). Suppose there exists $y$ with $f(y)=y$ (i.e. $d(y)=0$ )? Then $2x^2+2y^2\ge (x+d+y)^2$ . Choose $x$ in orbit close to $y$ … Since orbit points are spaced $d$ apart and go to infinity, but $y$ fixed — we can choose $m$ with $|x - y|\le d/2$ maybe (if orbit passes near $y$ ; orbit starts at $y_0$ and increases by $d$ , so for $y > y_0$ there is $x$ in orbit with $|x-y|\le d/2$ ; for large $y$ fine).

Check: $2x^2+2y^2 - (x+y+d)^2 \ge 0$ . With $x=y+\epsilon$ , $|\epsilon|\le d/2$ : $2(y+\epsilon)^2+2y^2 - (2y+\epsilon+d)^2 = 2y^2+4y\epsilon+2\epsilon^2+2y^2 - (4y^2 + 4y(\epsilon+d) + (\epsilon+d)^2) = 2\epsilon^2 - 4yd - (\epsilon+d)^2$ . For large $y$ this is very negative → contradiction! Great, provided we can find some $y$ with $d(y)$ bounded… but we need $d(y)$ small for large $y$ , not $=0$ . Let’s redo generally: we need for all $x$ in orbit and all $y$ : $2x^2+2f(y)^2 \ge (x+d+y)^2.$ With $f(y)=y+d(y)$ . Choose $y$ large, and $x$ in orbit with $x\approx y$ . Then LHS $\approx 2y^2+2(y+d(y))^2$ , RHS $\approx (2y+d+d(y))^2$ … Let’s compute with $x=y$ exactly (approx): $2y^2+2(y+e)^2 - (2y+d+e)^2$ where $e=d(y)$ : $=2y^2+2y^2+4ye+2e^2 - (4y^2+4y(d+e)+(d+e)^2) = 4ye+2e^2 - 4yd-4ye - (d+e)^2 = 2e^2 - 4yd - (d+e)^2$ . So need $2e^2 \ge 4yd + (d+e)^2$ , roughly $e^2 \gtrsim 4yd$ , i.e. $d(y) = e \gtrsim 2\sqrt{yd}$ . So $d(y)$ must grow like $\sqrt y$ at least. Then iterate: with larger $d$ ’s available, get stronger growth, contradiction. Let’s formalize.

So: Claim: if there exists $y_0$ with $d_0:=d(y_0)>0$ , then for all large $y$ , $d(y)\ge c\sqrt{y}$ roughly. Then take $y_1$ large with $d_1 = d(y_1)$ huge, and apply again: $d(y)\ge \sim 2\sqrt{y d_1}$ for large $y$ . Hmm, need care with “ $x\approx y$ ” error terms. Let’s do it cleanly.

Setup for contradiction argument. Suppose $d(y_0)=D>0$ . Orbit: $x_m = y_0+mD$ with $f(x_m)=x_m+D$ .

Constraint from (L) with $x=x_m$ , arbitrary $y$ : $2x_m^2 + 2f(y)^2 \ge (x_m + D + y)^2.\tag{B}$

Given $y > y_0$ , choose $m=\lceil (y-y_0)/D\rceil$ so that $0\le x_m - y< D$ . Write $x_m = y+\theta$ , $0\le \theta< D$ . Let $e=d(y)\ge0$ , $f(y)=y+e$ .

(B): $2(y+\theta)^2 + 2(y+e)^2 \ge (2y+\theta+D+e)^2$ . LHS $= 2y^2+4y\theta+2\theta^2+2y^2+4ye+2e^2$ . RHS $= 4y^2 + 4y(\theta+D+e) + (\theta+D+e)^2$ . So $4y\theta+4ye + 2\theta^2+2e^2 \ge 4y\theta+4yD+4ye + (\theta+D+e)^2$ ⟹ $2\theta^2 + 2e^2 \ge 4yD + (\theta+D+e)^2 \ge 4yD$ . Since $\theta < D$ : $2e^2 \ge 4yD - 2D^2$ , so $e^2 \ge 2yD - D^2$ .

Thus: $d(y)\ge \sqrt{2yD - D^2}$ for all $y>y_0$ . In particular for $y \ge D$ (and $y>y_0$ ), $d(y)\ge \sqrt{yD}$ .

Now pick $y_1 > \max(y_0, D)$ large; $D_1 := d(y_1)\ge \sqrt{y_1 D}$ . Apply the same claim with $y_0$ replaced by $y_1$ , $D$ by $D_1$ : for all $y > \max(y_1, D_1)$ : $d(y) \ge \sqrt{y D_1}$ . Hmm, we can iterate to get contradiction? Let’s instead find a direct contradiction: we need an upper bound on $d(y)$ .

Upper bound on $f$ : from (R)? (R) gives lower bounds on $f(x)$ . Hmm. Upper bound on $f(x)$ comes from (L): $f(x)\le \sqrt{2x^2+2f(y)^2} - y$ . Need a $y$ with $f(y)$ not too large relative to $y$ : but we just showed $f(y)$ is large for large $y$ … Use small $y$ ! For $y$ small (fixed), $f(y)$ is some value; then $f(x)\le \sqrt{2x^2+2f(y)^2}-y \le \sqrt2 x + \sqrt2 f(y) - y$ . So $f(x) \le \sqrt2 x + C$ where $C=\sqrt2 f(y_*) - y_*$ for a fixed $y_*$ . So $d(x)\le (\sqrt2 -1)x + C$ : linear growth bound. But our lower bound $d(y)\ge\sqrt{yD_1}$ with $D_1$ arbitrarily large… wait $D_1$ is fixed once $y_1$ chosen, and lower bound $\sqrt{yD_1}$ grows like $\sqrt y$ — weaker than linear. So need iteration to beat linear.

Iterate: from $d(y)\ge \sqrt{yD}$ for $y\ge \max(y_0,D)$ . Then choose $y_1$ : $D_1 = d(y_1)\ge \sqrt{y_1 D}$ . Then for $y\ge \max(y_1,D_1)$ : $d(y)\ge \sqrt{yD_1}\ge \sqrt{y\sqrt{y_1D}}$ . Hmm, to get contradiction with linear bound $d(y)\le \sqrt2 y + C$ : we need $\sqrt{y D_1} > \sqrt 2 y + C$ for some $y$ , i.e. $D_1 > 2y$ roughly, but constraint requires $y \ge D_1$ . Conflict: the bound requires $y>\max(y_1,D_1)$ , so $y> D_1$ , then $\sqrt{yD_1} < y$ . Hmm so no contradiction that way. Need better.

Let’s re-derive more carefully — maybe keep the full inequality: $2\theta^2+2e^2 \ge 4yD + (\theta+D+e)^2$ . Since $(\theta+D+e)^2 \ge e^2 + 2e(D+\theta)$ , we get $2\theta^2 + 2e^2 \ge 4yD + e^2 + 2e(D+\theta)$ , so $e^2 \ge 4yD + 2eD - 2\theta^2 + 2e\theta \ge 4yD - 2D^2$ (using $\theta<D$ , and $2eD+2e\theta\ge0$ ). Fine same.

Hmm. Alternatively, maybe use the fact that for each $y$ we have an orbit, and use two orbits with different $D$ s: apply (B) with $x$ from orbit of $y_0$ (spacing $D$ ) and $y$ from orbit of $y_1$ (spacing $D_1$ , $f(y)=y+D_1$ ). Then $e=D_1$ exactly. Condition: $2\theta^2 + 2D_1^2 \ge 4yD + (\theta+D+D_1)^2$ where $y$ ranges over $y_1 + kD_1$ (arbitrarily large!) and $\theta\in[0,D)$ . As $y\to\infty$ with $D_1$ fixed, RHS $\to\infty$ while LHS fixed → contradiction!!

Wait: but need $y$ in orbit of $y_1$ so $d(y)=D_1$ fixed, and $y\to\infty$ . Yes! Orbit of $y_1$ : $y_1+kD_1 \to\infty$ with $d = D_1$ constant. And $D>0$ fixed. Then inequality $2e^2 \ge 4yD - 2D^2$ with $e=D_1$ constant fails for large $y$ . Contradiction!

Even simpler: we don’t even need two orbits — take $y$ in the orbit of $y_0$ itself: $y = y_0+kD$ , $e = D$ , and $x=x_m$ with $x_m \approx y$ . But then $x$ and $y$ both in same orbit; the earlier computation showed same-orbit pairs are consistent… Let’s check: with $\theta$ : we need $x_m - y = \theta \in [0,D)$ ; but same orbit means $\theta$ is a multiple of $D$ , so $\theta = 0$ , i.e. $x=y$ . Then condition: $2e^2 \ge 4yD + (D+e)^2$ with $e=D$ : $2D^2 \ge 4yD + 4D^2$ : FALSE! Wait but earlier I checked same-orbit pairs satisfy (L). Let me recheck.

Earlier: $x=p$ , $f(x)=p+D$ , $y=q$ , $f(y)=q+D$ . (L): $2p^2+2(q+D)^2 \ge (p+D+q)^2$ . With $p=q=y$ : $2y^2+2(y+D)^2$ vs $(2y+D)^2$ : LHS $=4y^2+4yD+2D^2$ , RHS $=4y^2+4yD+D^2$ . LHS − RHS $= D^2>0$ ✓ fine. So my derived inequality must have a sign error. Let me redo.

Recompute with $x = y+\theta$ , $f(x)=x+D=y+\theta+D$ , $f(y)=y+e$ . (L): $2x^2+2f(y)^2 \ge (f(x)+y)^2$ . LHS $=2(y+\theta)^2+2(y+e)^2$ . RHS $=(y+\theta+D+y)^2 = (2y+\theta+D)^2$ .

Ah — I mistakenly included $e$ in RHS earlier. RHS shouldn’t have $e$ ! Correct RHS $=(2y+\theta+D)^2$ .

Redo: LHS $= 2y^2+4y\theta+2\theta^2 + 2y^2+4ye+2e^2 = 4y^2+4y\theta+4ye+2\theta^2+2e^2$ . RHS $= 4y^2 + 4y(\theta+D) + (\theta+D)^2$ . Inequality: $4ye + 2\theta^2+2e^2 \ge 4yD + (\theta+D)^2$ .

Check same-orbit $\theta=0,e=D$ : $4yD+2D^2 \ge 4yD+D^2$ ✓ consistent. Good.

So general: $4y(e - D) \ge (\theta+D)^2 - 2\theta^2 - 2e^2$ . Hmm. So if $e < D$ we get an upper bound on $y$ : i.e. $4y(D-e) \le 2\theta^2+2e^2 - (\theta+D)^2 \le 2\theta^2 + 2e^2 \le 2D^2+2e^2$ .

So: for $y>y_0$ with $d(y) = e < D$ : $4y(D-e) \le 2D^2+2e^2$ . Since $e<D$ : $\le 4D^2$ , so $y(D-e)\le D^2$ , giving $D - e \le D^2/y$ , i.e. $d(y) \ge D - \frac{D^2}{y}\quad\text{for all } y>y_0.$ (Holds also when $e\ge D$ trivially.)

So $\liminf_{y\to\infty} d(y)\ge D$ — but more importantly: apply this with an orbit point. Take $y$ in the orbit of $y_0$ ? gives $D \ge D - D^2/y$ fine.

Now use: pick $y_1$ with $D_1 = d(y_1) > D$ ? Do we know $d$ is unbounded? Let’s use bootstrapping: We know $d(y)\ge D - D^2/y$ for all $y>y_0$ . Choose $y_1 > 2y_0$ large, e.g. $y_1 \ge 2D^2/D$ ?? Let’s think: we want to produce a new orbit with larger spacing to iterate, contradiction with linear upper bound? Hmm, but $d(y)\ge D-D^2/y$ doesn’t give larger $D$ .

Alternatively use (R) to get an upper bound… Let’s fully exploit both inequalities for pairs.

Let’s parametrize $f(y)=y+d(y)$ , $d\ge0$ .

for general $x,y$ : $2x^2+2(y+d(y))^2 \ge (x+d(x)+y)^2$ . (R): $x+d(x)+y \ge 2\sqrt{x(y+d(y))}$ .

Set $x=y$ in (L): $2y^2+2(y+e)^2 \ge (2y+e')^2$ where $e=d(y)$ , $e'=d(y)$ — same! $x=y$ : $2y^2+2(y+e)^2 \ge (2y+e)^2$ : LHS $=4y^2+4ye+2e^2$ , RHS $=4y^2+4ye+e^2$ : true. no info.

Set $x = y+t$ , $t\in\mathbb R$ (as long as positive). (L): $2(y+t)^2 + 2(y+e)^2 \ge (2y+t+d(y+t))^2$ . Let $e_t = d(y+t)$ . Expand: $4y^2+4yt+2t^2+4ye+2e^2 \ge 4y^2+4y(t+e_t)+(t+e_t)^2$ , so $4y(e-e_t) + 2t^2+2e^2 \ge (t+e_t)^2.\tag{L'}$

So for all $y,t$ : $4y(d(y)-d(y+t)) \ge (t+d(y+t))^2 - 2t^2 - 2d(y)^2$ .

If $d(y+t) > d(y)$ (say $y+t$ has bigger defect), LHS negative — gives constraint: $4y(d(y+t)-d(y)) \le 2t^2+2d(y)^2 - (t+d(y+t))^2$ .

Now: Suppose $d$ is not identically 0. Pick $a$ with $d(a) = A>0$ . Orbit of $a$ : points $a+kA$ with defect $A$ . Now consider $y$ arbitrary with $d(y)=e$ small…

Take $y$ small-ish and $t$ such that $y+t = a+kA$ (large). Then $d(y+t) = A$ , $d(y)=e$ . Constraint (from L’): $4y(A - e) \le 2t^2 + 2e^2 - (t+A)^2$ . With $t = a+kA - y$ large, RHS $= 2t^2 - (t+A)^2 + 2e^2 = t^2 - 2tA - A^2 + 2e^2$ , which is large positive. No contradiction.

Instead choose $t$ small (i.e. $x$ near $y$ ) with $d$ differing a lot. We showed: for $x$ with $d(x)=D$ and $y$ near $x$ with $d(y)=e$ : constraint $4y(D-e)\le 2\theta^2+2e^2-(\theta+D)^2$ where $\theta = x-y$ . So if $e$ much smaller than $D$ and $y$ large → contradiction. So near any point of a big-defect orbit, defects must be nearly as big.

Given the orbit of $a$ with defect $A$ , points $a+kA$ are spaced $A$ ; for large $y$ , there’s an orbit point within distance $A$ above $y$ … So $d(y)\ge A - A^2/y$ for all $y>a$ (from earlier derivation with $D=A$ ). Similarly, can we get an upper bound on $d(y)$ from (L) with $y$ = orbit point and $x$ = general? Use (L’) form: $4y(d(y)-d(x)) \ge (t+d(x))^2 - 2t^2 - 2d(y)^2$ where $x=y+t$ . Take $y$ = orbit point of $a$ : $d(y)=A$ , and $x$ arbitrary near $y$ : $t$ small, $d(x)=D_x$ possibly large. Constraint: $4y(A - D_x)\ge (t+D_x)^2 - 2t^2 - 2A^2$ . If $D_x > A$ significantly and $y$ large: LHS very negative, RHS $\ge (t+D_x)^2-2t^2-2A^2$ . With $|t|\le$ small, RHS $\approx D_x^2 - 2A^2$ . So need $4y(A-D_x) \ge D_x^2 - 2A^2 - 2t^2$ , i.e. $4y(D_x - A) \le 2A^2+2t^2 - (t+D_x)^2$ . If $D_x>A$ , RHS $\le 2A^2 + 2t^2 - (t+D_x)^2$ . Need RHS $\ge 4y(D_x-A)>0$ .

So for $x$ near a large orbit point $y$ (within $|t|\le A$ ), $D_x$ can’t exceed $A$ by much: $4y(D_x-A) \le 2A^2+2t^2 \le 4A^2$ , so $D_x \le A + A^2/y$ .

So combining: for large $y$ , all defects near $y$ are $\approx A$ . So $d(x) \to A$ as $x\to\infty$ ?? Precisely: for $x$ large, $|d(x) - A|\le O(A^2/x)$ .

Hmm, so $d(x)\to A$ . Then $f(x) = x + A + o(1)$ . Does that contradict (R) or (L)? Let’s test whether $f(x)=x+A$ satisfies the original inequalities: (R): $f(x)+y = x+A+y \ge 2\sqrt{x f(y)} = 2\sqrt{x(y+A)}$ . Is $x+y+A \ge 2\sqrt{x(y+A)}$ ? Yes: AM-GM on $x$ and $y+A$ ✓. (L): $2x^2+2(y+A)^2 \ge (x+A+y)^2$ : with $u=x$ , $v=y+A$ : $2u^2+2v^2\ge (u+v)^2$ ✓. So $f(x)=x+A$ satisfies both inequalities!!

Uh oh. Let me verify original: $f(x)=x+c$ , $c>0$ . Left: $\sqrt{\frac{x^2+(y+c)^2}{2}} \ge \frac{x+c+y}{2}$ : QM ≥ AM for $x$ and $y+c$ ✓. Right: $\frac{x+c+y}{2}\ge \sqrt{x(y+c)}$ : AM ≥ GM ✓. So $f(x)=x+c$ works?! But then answer isn’t unique… but wait, we also need $f: \mathbb R_{>0}\to\mathbb R_{>0}$ , fine.

Hmm! But earlier we derived $f(f(y)) = 2f(y)-y$ : with $f(x)=x+c$ : $f(f(y)) = y+2c = 2(y+c) - y$ ✓ consistent.

Hmm, so the problem’s answer might be $f(x) = x$ only if additional constraints… Let me recheck the problem statement inequalities. The middle term: $\frac{f(x)+y}{2}$ . Left: $\sqrt{\frac{x^2+f(y)^2}{2}}$ . Right: $\sqrt{x f(y)}$ .

With $f(x)=x+c$ : LHS $=\sqrt{(x^2+(y+c)^2)/2}$ , middle $=(x+c+y)/2$ , right $=\sqrt{x(y+c)}$ . QM( $x$ , $y+c$ ) ≥ AM( $x,y+c$ ) ≥ GM. And middle $=(x+(y+c))/2$ = AM. Indeed exactly. So all $f(x)=x+c$ , $c\ge0$ work.

Hmm, so is the intended answer $f(x)=x+c$ ? Let me reconsider — perhaps this is a known competition problem. This resembles ISL 2020 A? Hmm. Let me recall: “Determine all functions $f:\mathbb R_{>0}\to\mathbb R_{>0}$ such that $\sqrt{\frac{x^2+f(y)^2}{2}}\ge \frac{f(x)+y}{2}$ …” Hmm, maybe original problem: $f(x)+y$ … Actually I recall a problem: find all $f$ with $\frac{f(x)+y}{2}\ge \sqrt{xf(y)}$ … Hmm.

Hmm wait, but maybe I should double check: is it $\sqrt{x f(y)}$ or $\sqrt{f(x) y}$ ? As given: $\sqrt{xf(y)}$ . Fine.

But hold on: with $f(x)=x+c$ , the two inequalities both become QM-AM-GM for the pair $(x, y+c)$ , requiring middle $= \frac{x+(y+c)}{2}$ , and indeed $\frac{f(x)+y}{2} = \frac{x+c+y}{2}$ ✓. So yes, works.

Hmm, but then the answer: all $f(x)=x+c$ with $c\ge0$ ? Let’s double-check whether $c<0$ possible: $f(x)=x+c$ with $c<0$ maps to positive? Not for small $x$ . And we proved $d\ge0$ . Let’s verify our proof that $d$ is constant and thus $f(x)=x+c$ .

Let’s continue the derivation rigorously to conclude $d$ constant.

We have $d(y)\ge 0$ and $f(f(y))=2f(y)-y$ , orbits are APs with constant defect.

Key inequality (L’): for all $x,y>0$ : $4y\,(d(y)-d(x)) + 2(x-y)^2 + 2d(y)^2 \ \ge\ (x-y+d(x))^2. \tag{L'}$ Let me re-derive carefully. (L): $2x^2 + 2f(y)^2 \ge (f(x)+y)^2$ . With $f(x)=x+d(x)$ , $f(y)=y+d(y)$ , $t = x-y$ : LHS $=2(y+t)^2 + 2(y+d(y))^2 = 2y^2+4yt+2t^2+2y^2+4y\,d(y)+2d(y)^2$ . RHS $=(y+t+d(x)+y)^2 = (2y + t + d(x))^2 = 4y^2 + 4y(t+d(x)) + (t+d(x))^2$ . LHS−RHS $= 4yt+2t^2+4y d(y)+2d(y)^2 - 4yt - 4y d(x) - (t+d(x))^2 = 4y(d(y)-d(x)) + 2t^2+2d(y)^2 - (t+d(x))^2 \ge 0$ ✓ matches (L’).

Now to prove $d$ constant. Suppose $d(x_0) = D$ for some $x_0$ . Orbit: $x_k = x_0 + kD$ with $d(x_k)=D$ , for all $k\ge 0$ (if $D>0$ these tend to infinity; if $D=0$ orbit is a single point).

Case: suppose there exist $a,b$ with $d(a) \ne d(b)$ . WLOG… let’s handle via:

Claim 1: For all $x, y$ with $d(x) > d(y)$ : $4y(d(x)-d(y)) \le 2(x-y)^2 + 2d(y)^2 - (x-y+d(x))^2$ . That’s just (L’) rearranged. Hmm, RHS can be big if $|x-y|$ big. So constraint is only strong when $x$ near $y$ .

Strategy: Show $d$ is constant.

Suppose $d(a)=A$ , $d(b)=B$ with $A\neq B$ ; WLOG $A > B \ge 0$ … hmm we need to compare defects at nearby large arguments. Use orbits: orbit of $a$ has points $a+kA$ (spacing $A$ , if $A>0$ ), orbit of $b$ has points $b+kB$ (spacing $B$ ; if $B=0$ then just $b$ ).

Take $y = b+kB$ (large, if $B>0$ ) and $x = a+mA$ chosen with $x-y = t \in [0,A)$ . Then $d(y)=B$ , $d(x)=A$ , and (L’) gives: $4y(B-A) + 2t^2 + 2B^2 \ge (t+A)^2$ . LHS $\le 4y(B-A) + 2A^2+2B^2$ , and RHS $\ge A^2$ . So need $4y(A-B) \le 2A^2+2B^2 - A^2 \le 2A^2 + 2B^2$ . Since $y$ can be arbitrarily large (as $B>0$ ), contradiction.

If $B=0$ : orbit of $b$ is just $\{b\}$ ; then we can’t take $y$ large. But then use: $d(y)\ge$ something? Alternative: If $B = 0$ , i.e. $f(b)=b$ , use $y=b$ fixed and $x$ from orbit of $a$ with $t = x - b \in [0,A)$ : $4b(0-A) + 2t^2 + 0 \ge (t+A)^2$ ⟹ $2t^2 \ge (t+A)^2 + 4bA \ge A^2 + 4bA$ . But $t<A$ so $2t^2 < 2A^2$ . Need $2A^2 > A^2+4bA$ ⟹ $A > 4b$ . Hmm, not immediately contradictory: if $A$ large relative to $b$ , no contradiction from this pair alone. But we can also choose $t$ negative? $x$ from orbit with $x < b$ ? Only if $a < b$ . Hmm.

Alternative cleaner approach: Show $d$ is constant by showing $d(y)$ determined: use (L’) with $x$ ranging over the orbit of $y$ itself… no.

Let’s think again: we want a proof that $d$ is constant. Better approach: derive from (L’) an inequality forcing $d(x)\le$ something like $d(y) + $ small when $x$ near $y$ and $y$ large; and combined with orbits reaching infinity.

Suppose $d(a) = A > 0$ (some point). Orbit $a+kA\to\infty$ with defect $A$ .

Step: show for every $y>0$ , $d(y) \ge A - A^2/y$ … derived earlier: take $x$ in orbit of $a$ with $t=x-y\in[0,A)$ (possible if $y > a$ ? need $a+mA \ge y$ and $< y+A$ : exists for any $y\ge a$ ; for $y<a$ choose $m=0$ ? then $t = a-y$ could exceed $A$ . So restrict to $y\ge a$ .) Then (L’): $4y(d(y)-A) + 2t^2 + 2d(y)^2 \ge (t+A)^2$ . Since $(t+A)^2 \ge 2t^2$ is false in general… let’s bound: $2t^2 - (t+A)^2 = t^2 - 2tA - A^2 \le -A^2 <0$ for $t\in[0,A)$ ? At $t=A$ : $A^2-2A^2-A^2=-2A^2$ ; at $t=0$ : $-A^2$ . So $2t^2-(t+A)^2 \le -A^2$ . Then (L’) gives $4y(d(y)-A) + 2d(y)^2 \ge A^2$ . So if $d(y) < A$ : $2d(y)^2 \ge A^2 + 4y(A-d(y)) \ge 4y(A-d(y))$ , and $d(y)<A$ ⟹ $2A^2 > 4y(A-d(y))$ ⟹ $A - d(y) < \frac{A^2}{2y}$ . So $d(y) > A - \frac{A^2}{2y}\quad \text{for all } y\ge a. \tag{C}$

Also, upper bound: take $y$ in orbit of $a$ (so $d(y)=A$ , $y=a+kA$ arbitrarily large), and $x$ arbitrary. (L’): $4y(A - d(x)) + 2t^2+2A^2 \ge (t+d(x))^2$ where $t = x-y$ . For fixed $x$ , choose $k$ large so $y\to\infty$ : then if $d(x) > A$ , LHS $\to -\infty$ while RHS… but $t = x-y \to -\infty$ so $2t^2$ grows like $2y^2$ and $(t+d(x))^2 \approx y^2$ . Hmm: LHS $\approx 4y(A-d(x)) + 2y^2$ , RHS $\approx y^2$ . So LHS − RHS $\approx y^2 > 0$ . No contradiction. Need $x$ near $y$ : but $x$ arbitrary fixed; we need for each large $y$ an $x$ near it with large defect. That’s the issue: to bound defects above we need…

Alternative: use (R) for upper bound on $d$ ! (R): $f(x)+y \ge 2\sqrt{xf(y)}$ : $x+d(x)+y \ge 2\sqrt{x(y+d(y))}$ . Choose $y$ so that RHS large: pick $y$ with $y + d(y) = x$ ideally (AM-GM equality): then RHS $=2x$ , so $d(x) \ge x - y = d(y)$ . Hmm that gives lower bound again. Hmm (R) gives lower bounds on $d(x)$ generally. Indeed (R): $d(x) \ge 2\sqrt{x(y+d(y))} - x - y$ .

Upper bounds on $d$ must come from (L). (L): $2x^2 + 2(y+d(y))^2 \ge (x+d(x)+y)^2$ , i.e. $d(x) \le \sqrt{2x^2+2f(y)^2} - x - y$ . To make RHS small, want $f(y)$ small and $y$ … minimize over $y$ : RHS as function with $f(y)\ge y$ : $\ge \sqrt{2x^2+2y^2}-x-y \ge 2x - x - ...$ : at $y = x$ : $2x - x - x = 0$ ! So if there’s $y$ with $f(y)=y$ (defect 0) we’d get $d(x)\le \sqrt{2x^2+2y^2}-x-y$ which at $y=x$ gives 0. But defect-0 points: only if $d\equiv$ ? We know from (C) that if some $A>0$ exists, then $d(y) > A - A^2/(2y) >0$ for $y > A/2$ ; so defect-0 points are bounded: $d(y)=0 \Rightarrow y \le A/2$ .

Let’s get the upper bound using (C) plus (L). Actually here’s a cleaner plan: Show that $d$ is bounded above (or find the exact structure).

Let $S = \sup_y d(y) \in (0,\infty]$ (assuming not identically 0; if $d\equiv 0$ then $f(x)=x$ , a special case of $x+c$ ).

Suppose $d(a) = A > 0$ . By (C), $\liminf_{y\to\infty} d(y) \ge A$ . Take any $B$ with $B = d(b) > A$ for some $b$ — then by (C) applied with $B$ : $\liminf d \ge B$ . Iterating: $\liminf_{y\to\infty} d(y) \ge S$ . Also need: is $S$ finite? Use (L) with a fixed pair to bound $d(x)$ linearly: $d(x)\le \sqrt{2x^2+2f(y_*)^2} - x - y_* \le (\sqrt2-1)x + \sqrt2 f(y_*) - y_*$ . So $d(x) = O(x)$ , but $S$ could be $\infty$ via unbounded $d$ .

Hmm, let’s aim differently: show directly that $d(x) d(y) + $ tiny for large…

New idea: use (L’) with $y$ large in orbit and $x$ = orbit point of another orbit near it, as I did—the contradiction worked when both orbits are infinite (both defects positive). The only problematic case was comparing a positive-defect orbit with a zero-defect point. But actually we might avoid: Suppose $d$ takes two distinct positive values $A>B>0$ at $a, b$ . Then contradiction as shown. So $d$ takes at most one positive value, plus possibly the value 0. So $d(y)\in\{0,A\}$ for all $y$ (with $A>0$ if $d\not\equiv0$ ). And by (C), $d(y)=A$ for all $y>A/2$ . Remains: points $y\le A/2$ with $d(y)=0$ , i.e. $f(y)=y$ . But then check (L) or (R) with such $y$ : $f(y)=y$ , $y \le A/2$ . Take $x$ large with $d(x)=A$ : (L’): $4y(0 - A) + 2t^2 + 0 \ge (t+A)^2$ where $t = x-y$ large positive: $2t^2 - (t+A)^2 = t^2-2tA-A^2 \ge 4yA$ ? For large $t$ this holds. Hmm so no contradiction from large $x$ . Take $x$ near $y$ : $x$ with $d(x)=A$ and $x$ close to $y$ : need such $x$ ; all $x>A/2$ have $d(x)=A$ . And $y\le A/2$ . So closest is $x$ slightly $> A/2$ , i.e. $t = x - y \ge A/2 - y$ … hmm we need $t$ small-ish. Condition needed: $2t^2 - (t+A)^2 \ge 4yA$ , i.e. $t^2 - 2At - A^2 - 4yA\ge0$ ⟹ $t \ge A + \sqrt{2A^2+4yA}$ (positive root: $t = A+\sqrt{A^2+A^2+4yA}$ ). So for $x$ with $t = x-y < A+\sqrt{2A^2+4Ay}$ and $d(x)=A$ we get contradiction. Since all $x > A/2$ have defect $A$ , choose $x = A/2 + \epsilon$ … wait need also check whether $x$ could have $d(x)=0$ ; we established $d(x)=A$ for $x>A/2$ . Take $x = A$ (say, $A > A/2$ ✓ so $d(A)=A$ ). Then $t = A - y \ge A - A/2 = A/2 >0$ , and is $t < A + \sqrt{2A^2+4Ay}$ ? $t = A-y < A < A+\sqrt{\cdot}$ ✓. So contradiction. Hence no $y$ with $d(y)=0$ when $A>0$ . Wait but careful: also need to double check the case $d\equiv A$ everywhere including small $y$ — that’s $f(x)=x+A$ which is fine.

Also need to handle: what if $d$ takes value 0 at some points and $d\not\equiv 0$ : handled above → contradiction. Actually wait: I used “ $d$ takes at most one positive value” which came from the two-positive-orbit contradiction; need to also verify that argument’s details (choosing $x$ in orbit $A$ with $x-y\in[0,A)$ where $y$ in orbit $B$ arbitrarily large). Let’s re-verify: orbit of $a$ : $a+mA$ , $m\ge0$ , defect $A$ . Orbit of $b$ : $b+kB$ , defect $B$ . Given large $y = b+kB$ , choose $m = \lceil (y-a)/A\rceil$ so $x = a+mA \in [y, y+A)$ , so $t = x-y \in [0,A)$ . (L’): $4y(d(y)-d(x)) + 2t^2 + 2d(y)^2 \ge (t+d(x))^2$ , i.e. $4y(B-A) + 2t^2+2B^2 \ge (t+A)^2 \ge 0$ . Since $A>B$ : LHS $\le 4y(B-A) + 2A^2 + 2B^2 \to -\infty$ as $y\to\infty$ . Contradiction ✓. Good, this requires $B>0$ to have $y\to\infty$ ; and requires $A > B$ ; we just need two distinct positive values. If $A>B>0$ then contradiction. Great.

Hmm, but actually simpler: we can unify. Let me restructure the whole proof cleanly:

Proof structure: 1. From (L) with $x=f(y)$ and (R) with $x=f(y)$ : get $f(f(y))=2f(y)-y$ . 2. Iterating: orbit is AP ⟹ $f(y)\ge y$ ; define $d(y)=f(y)-y\ge0$ , and $f(y+kd(y)) = y+(k+1)d(y)$ , i.e., $d(y+kd(y))=d(y)$ for all $k\ge0$ . 3. Master inequality (L’) from (L). 4. Show $d$ constant: (a) If two distinct positive values $A>B>0$ → contradiction via (L’) with $y\to\infty$ along $B$ -orbit and $x$ in $A$ -orbit within $[y,y+A)$ . (b) So $d(\mathbb R_{>0}) \subseteq \{0, A\}$ for a single $A\ge0$ . If $A=0$ , $f(x)=x$ . If $A>0$ : show $d(y)=0$ impossible: from (C) [derived via (L’) with $A$ -orbit], $d(y)>A - A^2/(2y)$ for $y \ge a$ ; hmm, that requires $y\ge a$ where $a$ is a base point. Actually with the dichotomy $d\in\{0,A\}$ , (C) gives $d(y)=A$ for all $y\ge \max(a, A/2)$ … then for the remaining $y$ with $d(y)=0$ , use (L’) with a suitable $x$ having $d(x)=A$ and get contradiction. Let’s make this clean.

Alternatively for step (b), cleaner: suppose $d(y_1)=0$ and $d(x_1)=A>0$ . Derive contradiction directly from (L’) without needing (C)? (L’) with $y=y_1$ , $x$ in orbit of $x_1$ : $-4y_1 A + 2t^2 + 0 \ge (t+A)^2$ where $t = x - y_1$ , $x = x_1 + mA$ , need $t$ small: $t \ge x_1 - y_1$ can’t control if $x_1 \gg y_1$ . Hmm. But we can also use orbit of $y_1$ ? It’s trivial ( $d=0$ means $f(y_1)=y_1$ fixed point).

Alternatively, use (R) to derive a contradiction with a fixed point: If $f(y_1)=y_1$ , (R): $f(x)+y_1 \ge 2\sqrt{x y_1}$ — no. (L): $2x^2+2y_1^2 \ge (f(x)+y_1)^2$ . So $f(x) \le \sqrt{2x^2+2y_1^2} - y_1$ . Take $x=y_1$ : $f(y_1)\le 2y_1 - y_1 = y_1$ ✓ nothing. Take $x$ general: $d(x) \le \sqrt{2x^2+2y_1^2} - x - y_1$ . Note this equals $\frac{2x^2+2y_1^2 - (x+y_1)^2}{\sqrt{2x^2+2y_1^2}+x+y_1} = \frac{(x-y_1)^2}{\sqrt{2x^2+2y_1^2}+x+y_1}$ . So $d(x) \le \frac{(x-y_1)^2}{\sqrt{2x^2+2y_1^2}+x+y_1}\le \frac{(x-y_1)^2}{2(x+y_1)}.$ Nice clean bound! So a fixed point $y_1$ forces $d(x)$ small for $x$ near $y_1$ . In particular $d$ is continuous-ish at fixed points. Now if $d(x_1) = A>0$ : then $d(x)=A$ for all $x$ in orbit $x_1+mA$ . Take such $x$ far from… hmm we need $x$ near $y_1$ . Orbit points are $\ge x_1$ and spaced $A$ . Bound: $A = d(x) \le \frac{(x-y_1)^2}{2(x+y_1)}$ for all orbit points $x$ . As $x\to\infty$ RHS $\to\infty$ ; not contradiction. Need orbit point near $y_1$ , i.e. $x - y_1$ small compared to $\sqrt{A(x+y_1)}$ … hmm.

Use combination: from (C) (derived from $A$ -orbit), $d(y)>A - A^2/(2y)$ for $y\ge x_1$ . Also from fixed point $y_1$ bound: $d(y)\le \frac{(y-y_1)^2}{2(y+y_1)}$ . For $y$ slightly bigger than $y_1$ ?? But (C) only applies for $y \ge x_1$ (since we need an orbit point in $[y,y+A)$ ; actually for any $y \ge x_1$ we can find orbit point in $[y, y+A)$ ). Hmm, so we need to compare at large $y$ : (C): $d(y) \ge A - A^2/(2y)\ge A/2$ for $y\ge A$ . Fixed point bound at large $y$ gives $d(y)\le (y-y_1)^2/(2(y+y_1))\approx y/2$ — weaker. No contradiction. Hmm!

So need a better approach for case (b). Wait — but maybe case (b) is genuinely impossible only through subtler reasoning; let’s think about whether a function with $d\in\{0,A\}$ , $d=A$ on $(A/2,\infty)$ … but wait, we need $d(y+kd(y))=d(y)$ : for $y$ with $d(y)=0$ fine (fixed point). Consider candidate: $f(y)=y$ for $y\le c$ and $f(y)=y+A$ for $y>c$ . Check consistency: for $y>c$ , orbit: $y+A > c$ ✓ defect $A$ ✓. Need to check (L) and (R) for all pairs.

(R): $f(x)+y\ge 2\sqrt{xf(y)}$ . Take $x\le c$ (so $f(x)=x$ ) and $y>c$ (so $f(y)=y+A$ ): need $x+y\ge 2\sqrt{x(y+A)}$ , i.e. $(x+y)^2 \ge 4xy+4xA$ , i.e. $(x-y)^2\ge 4xA$ . Take $x=c$ , $y$ slightly $>c$ : LHS $\approx0$ , RHS $=4cA>0$ : FALSE. So violated ✓ contradiction as expected. So (R) kills it! I should use (R) too for step (b). Indeed use (R) with $x$ a fixed point ( $d(x)=0$ ) and $y$ with $d(y)=A$ , $y$ near $x$ : (R): $x + y \ge 2\sqrt{x(y+A)}$ ⟹ $(x-y)^2 \ge 4xA$ .

So: if $d(x)=0$ and $d(y)=A$ then $(y-x)^2 \ge 4xA$ . Hmm, need $y$ near $x$ : given $x$ fixed point, and $A>0$ with $A$ -orbit points $\to\infty$ , and by (C) $d(y)=A$ for all large $y$ … $(y-x)^2\ge 4xA$ holds for large $y$ . Not contradiction. Need $y$ with $d(y)=A$ and $y$ close to $x$ (within $2\sqrt{xA}$ ).

Hmm. So we need to establish that defects are “locally constant”/spread everywhere. Let’s get a stronger tool: (C) says $d(y)\ge A - A^2/(2y)$ for $y\ge a$ (base point of the $A$ -orbit). If also $d$ only takes values in $\{0,A\}$ , then $d(y)=A$ for all $y \ge \max(a, A/2)$ … but is that enough to find $y$ with $d(y)=A$ near a fixed point $x$ ? Fixed point $x$ could be small, e.g. $x < A/2$ and $a$ large. Then we need $y$ with $d(y)=A$ within distance $2\sqrt{xA}$ of $x$ . Points with $d=A$ are all $\ge \max(a,A/2)$ -ish, so distance $\ge \max(a,A/2) - x$ . Need $\max(a,A/2)-x < 2\sqrt{xA}$ ? Not guaranteed.

Hmm, so we need to push (C) down to small $y$ or find better bounds. Let’s reconsider: maybe use (R) to propagate defects downward.

Let’s use (R) generally: $x + d(x) + y \ge 2\sqrt{x(y+d(y))}$ for all $x,y$ . Set $y$ such that $y + d(y)$ is large relative… Let’s set $y = x - d(x)$ ?? Need positive. Hmm. Let’s think of (R) as: for all $x,y$ : $\left(\sqrt{x} - \sqrt{y+d(y)}\right)^2 \ge$ … compute: $x+y+d(x) - 2\sqrt{x(y+d(y))} \ge 0$ . Write $P = x$ , $Q = y+d(y) = f(y)$ . Then $P + Q - 2\sqrt{PQ} + d(x) - d(y) \ge 0$ , i.e. $(\sqrt{x}-\sqrt{f(y)})^2 \ge d(y) - d(x).\tag{R'}$ Similarly (L): $2x^2+2f(y)^2 \ge (f(x)+y)^2 = (x + d(x) + y)^2 = (x + f(y) + d(x) - d(y))^2$ . Let $u = x$ , $v=f(y)$ , $\delta = d(x)-d(y)$ : $2u^2+2v^2 \ge (u+v+\delta)^2$ . So $(u-v)^2 \ge 2\delta(u+v) + \delta^2 = (u+v+\delta)^2 - (u+v)^2 ...$ let me verify: $2u^2+2v^2 - (u+v)^2 = (u-v)^2$ . And $(u+v+\delta)^2 = (u+v)^2 + 2\delta(u+v)+\delta^2$ . So (L) ⟺ $(u-v)^2 \ge 2\delta(u+v) + \delta^2$ , i.e. $(x - f(y))^2 \ \ge\ 2(d(x)-d(y))(x+f(y)) + (d(x)-d(y))^2.\tag{L''}$ And (R’): $(\sqrt x - \sqrt{f(y)})^2 \ge d(y)-d(x)$ .

These are nice. Note in (L’’), if $d(x) > d(y)$ then $(x-f(y))^2 \ge 2(d(x)-d(y))(x+f(y))$ , strong constraint (need $x$ far from $f(y)$ ).

Let $\delta = d(x)-d(y)$ . - If $\delta>0$ : (L’‘) gives $|x - f(y)| \ge$ roughly $\sqrt{2\delta(x+f(y))}$ . - If $\delta<0$ : (R’) gives $|\sqrt x - \sqrt{f(y)}| \ge \sqrt{-\delta}$ , i.e. $|x-f(y)| \ge \sqrt{-\delta}(\sqrt x + \sqrt{f(y)})$ , i.e. $\ge$ roughly $\sqrt{-\delta \cdot (x+f(y))}$ too. Nice symmetry!

So: for all $x,y$ : $|x - f(y)|$ is bounded below in terms of $|d(x)-d(y)|$ . Both give: if $d(x)\ne d(y)$ then $x$ and $f(y)$ are separated.

Now, the plan: given any $y$ , take $x = f(y)$ : then $\delta = d(f(y)) - d(y) = 0$ (consistent, we know).

To show $d$ constant: given $y_1,y_2$ with $d(y_1)\ne d(y_2)$ , we want contradiction. Consider $x$ ranging over the orbit of $y_1$ ( $d = d(y_1)$ on it) and $f(y)$ over orbit-ish of $y_2$ . Hmm the previous argument: with $A = d(y_1) > B = d(y_2) \ge 0$ : Use (R’) with $x$ in orbit of $y_2$ ?? Let’s redo with these clean forms.

Case $B>0$ : orbit of $y_2$ : $f$ values $y_2 + kB$ , all with $d=B$ ; orbit of $y_1$ : points $y_1+mA$ with $d = A$ . Take $y = y_2+kB$ (so $f(y) = y_2+(k+1)B$ , $d(y)=B$ ) and $x = y_1+mA$ with $d(x)=A > B$ , so $\delta = A - B>0$ . (L’’): $(x - f(y))^2 \ge 2(A-B)(x+f(y))$ . Choose $m$ so that $|x - f(y)| < A$ (possible: orbit points spaced $A$ , and $f(y)$ can be huge). Then $A^2 > 2(A-B)(x+f(y)) \ge 2(A-B)f(y) \to \infty$ . Contradiction ✓ (need $f(y)\to\infty$ ✓ since $B>0$ ).

Case $B = 0$ ( $f(y_2)=y_2$ , fixed point). Now use (R’) with $y=y_2$ : $(\sqrt x - \sqrt{y_2})^2 \ge d(y_2) - d(x) = -d(x)$ : trivially true. Use (L’’) with $y=y_2$ , $x$ arbitrary: $(x-y_2)^2 \ge 2 d(x)(x+y_2) + d(x)^2$ . So $d(x) \le \frac{(x-y_2)^2}{2(x+y_2)}$ roughly (as computed). Now take $x$ = orbit point of $y_1$ : $d(x)=A$ , $x = y_1+mA$ . Then $A \le \frac{(x-y_2)^2}{2(x+y_2)}$ . For large $x$ that’s satisfied. So still stuck: need small $x$ with defect $A$ or something. Hmm.

We need to also use (R’) to force defects to be large at small arguments? (R’): $(\sqrt x - \sqrt{f(y)})^2 \ge d(y)-d(x)$ , i.e. $d(x) \ge d(y) - (\sqrt x - \sqrt{f(y)})^2$ . So defect at $x$ is bounded below by $d(y)$ minus separation. Take $y$ in the $A$ -orbit: $f(y)= y_1+(m+1)A$ can be any orbit point (large). To get a good lower bound on $d(x)$ for small $x$ , need $f(y)$ close to $x$ — but orbit points are large. Hmm. So with a fixed point $y_2$ small and an $A$ -orbit starting large, the constraints between them: $(x-y_2)^2 \ge 2A(x+y_2)+A^2$ for all $x$ in $A$ -orbit; and separately need consistency for all other points.

But wait: we haven’t used the full strength: for every $y>0$ there’s an orbit; every point lies in its own orbit. The set of defect values: suppose $d$ takes values $A>0$ and $0$ . All points $x$ with $d(x)=A$ satisfy $(x-y_2)^2 \ge 2A(x+y_2)+A^2$ where $y_2$ any fixed point. Also, consider a point $z$ with $d(z)=A$ ; then, is there a constraint forcing points just below $z$ to have defect $A$ too (hence eventually reaching near $y_2$ )? Consider $x$ with $d(x)=0$ (fixed point) and compare with $y=z$ : $\delta = d(x)-d(z) = -A <0$ , so (R’): $(\sqrt x - \sqrt{f(z)})^2 \ge A$ where $f(z) = z+A$ . So $|\sqrt x - \sqrt{z+A}|\ge \sqrt A$ for every fixed point $x$ and every $A$ -point $z$ . Also (L’’) with $x$ fixed point ( $d(x)=0$ ), $y=z$ ( $d=A$ ): $\delta = -A$ : $(x - (z+A))^2 \ge -2A(x+z+A) + A^2$ , weaker.

So constraints: for all fixed points $p$ (defect 0) and all $A$ -points $z$ : (i) from (L’‘) with $x=z$ (defect $A$ ), $y=p$ (defect 0, $f(p)=p$ ): $(z-p)^2 \ge 2A(z+p)+A^2$ . (ii) from (R’) with $x=p$ , $y=z$ : $(\sqrt p - \sqrt{z+A})^2 \ge A$ .

Are these satisfiable? E.g. $A$ -points all huge and $p$ small: (i): $z^2 \gtrsim 2Az$ → $z \gtrsim 2A$ fine. (ii): $\sqrt{z+A}-\sqrt p \ge \sqrt A$ fine for large $z$ . So need more constraints to rule out. Also constraints among the structure: every point has a defect in $\{0,A\}$ ; the set $Z$ of $A$ -points must be closed under $z\mapsto z+A$ , and set $P$ of fixed points; $P\cup Z = \mathbb R_{>0}$ .

But (C)-type: by (i), all $A$ -points $z$ satisfy $z \ge$ something like $2A + \dots$ , and by (C) all $y \ge \max(\text{base}, ...)$ are $A$ -points. Actually (C) argument: for any $y$ , choose $A$ -point $x$ with $x - y\in[0,A)$ (exists provided $y$ is at most the max… since $A$ -points include an AP going to infinity, for any $y \ge z_0$ (some $A$ -point) there’s an $A$ -point in $[y,y+A)$ ). Then (L’’) with this $x$ and $y$ : $\delta = A - d(y)$ . If $d(y) = 0$ : $(x-y)^2 \ge 2A(x+y)+A^2$ ⟹ $A^2 > 2A(x+y) \ge 2A\cdot y$ ⟹ $y < A/2$ . And $x < y+A < 3A/2$ , but also (i) requires $(x-y)^2 \ge 2A(x+y) + A^2 > A^2$ , so $x-y > A$ — contradicts $x-y<A$ !

So: for any $y \ge z_0$ (where $z_0$ is the smallest available $A$ -orbit base), $d(y) \ne 0$ , so $d(y)=A$ (in the dichotomy). But actually we get more directly: for any $y\ge z_0$ , choosing $x$ in $[y, y+A)$ with $d(x)=A$ , (L’’) gives $(x-y)^2 \ge 2(A - d(y))(x+y) + (A-d(y))^2$ . If $d(y) < A$ : LHS $< A^2$ , RHS $\ge 2(A-d(y))\cdot 2y$ . So $A - d(y) < \frac{A^2}{4y}$ — that’s (C) refined. Fine. Combined with dichotomy $d\in\{0,A\}$ : for $y \ge z_0$ with $y \ge A/4$ … get $d(y)=A$ .

Hmm, but the issue remains: fixed points $p < z_0$ . We need to rule out $p$ small with all $A$ -points $\ge z_0$ where $z_0$ might be large. But wait: is $z_0$ arbitrary? The set of $A$ -points: contains $y_1$ and its orbit, and contains all $y\ge \max(y_1, A/4)$ hmm. But could there be an $A$ -point smaller? Let’s think about whether the configuration ( $P = (0,c]$ , $Z = (c,\infty)$ with $A$ -defect) can satisfy all constraints; earlier I checked with (R) that it fails: with $x=c\in P$ ( $f(x)=x=c$ ) and $y$ slightly above $c$ ( $f(y)=y+A$ ): (R’): $(\sqrt c - \sqrt{y+A})^2 \ge d(y)-d(x) = A$ . With $y\downarrow c$ : $(\sqrt{c+A}-\sqrt c)^2 \ge A$ ? Is that possible? $(\sqrt{c+A}-\sqrt{c})^2 = 2c + A - 2\sqrt{c(c+A)}$ . Compare with $A$ : need $2c - 2\sqrt{c(c+A)} \ge 0$ ⟺ $c \ge \sqrt{c(c+A)}$ ⟺ $c^2 \ge c^2 + cA$ : false. So contradiction ✓ (as long as we can take $y$ arbitrarily close to $c$ from above; with $P=(0,c]$ , $Z=(c,\infty)$ ).

But in general $P$ and $Z$ partition $(0,\infty)$ with $P$ bounded (contained in $(0, A/4 + \dots)$ or below $z_0$ …). Hmm. We need: exists $p\in P$ and $z\in Z$ with $\sqrt{z+A} - \sqrt p < \sqrt A$ , i.e. $z + A < (\sqrt p + \sqrt A)^2 = p + A + 2\sqrt{pA}$ , i.e. $z < p + 2\sqrt{pA}$ . So need an $A$ -point $z$ less than $p + 2\sqrt{pA}$ where $p$ is a fixed point. Equivalently, since $P$ is bounded above (by, say, $M$ ) and $Z \supseteq (M,\infty)$ : take $p = \sup P$ (or close to it, $p \in P$ with $p > \sup P - \epsilon$ ), then $z$ slightly bigger than $\sup P$ : $z \in Z$ can be chosen with $z < \sup P + \epsilon'$ . Then need $\sup P + \epsilon' < p + 2\sqrt{pA}$ : since $p > \sup P - \epsilon$ , suffices $\epsilon+\epsilon' < 2\sqrt{pA}$ ✓ for small $\epsilon,\epsilon'$ (with $p>0$ bounded away from 0 — need $\sup P>0$ ✓ since $P$ nonempty and contains some $p_0>0$ , so $\sup P \ge p_0>0$ ; choose $\epsilon < p_0/2$ so $p \ge p_0/2$ … careful: we need $p\in P$ with $p$ near $\sup P$ ; fine).

Also need $z\in Z$ with $z$ slightly above $\sup P$ : since $Z \supseteq (\sup P, \infty)$ (as points $> \sup P$ aren’t in $P$ ), yes ✓ — since $P\cup Z=(0,\infty)$ and everything above $\sup P$ is not in $P$ , hence in $Z$ .

Great, so the case- $B=0$ contradiction works via (R’). And actually we can streamline: we don’t need the dichotomy first — we could directly define things generally. Let’s see if there’s a cleaner unified route.

Cleaner route attempt. Let $d(y)=f(y)-y \ge 0$ . Suppose $d$ is not constant: exist $y_1,y_2$ with $d(y_1) > d(y_2) \ge 0$ . Set $A = d(y_1)>0$ , $B=d(y_2)$ .

Key relations: (L’‘): $(x-f(y))^2 \ge 2(d(x)-d(y))(x+f(y)) + (d(x)-d(y))^2$ . (R’’): $(\sqrt x-\sqrt{f(y)})^2 \ge d(y)-d(x)$ .

Case 1: $B>0$ : as shown contradiction via (L’‘) with $y$ in $B$ -orbit going to infinity and $x$ in $A$ -orbit close to $f(y)$ . Actually let’s double check: we need $x$ in $A$ -orbit with $|x - f(y)| < A$ : yes since $A$ -orbit is an AP with difference $A$ tending to $\infty$ , and $f(y)$ large. Then (L’’): $A^2 > (x-f(y))^2 \ge 2(A-B)(x+f(y)) \ge 2(A-B)f(y)\to\infty$ . Contradiction ✓.

Case 2: $B=0$ . Then set $P=\{p: d(p)=0\}\ne\emptyset$ , and from Case-1-type reasoning, all positive defect values are equal? Hmm, in Case 2 we might have multiple positive values, but then Case 1 applies. So assume (for the structure) that $d$ takes exactly the values $\{0, A\}$ with $A>0$ . Then proceed as above: - $P$ bounded above: from (L’‘) with $p\in P$ , $x$ an $A$ -point: $(x-p)^2 \ge 2A(x+p)+A^2$ . Hmm this bounds… For fixed $A$ -point $x$ , gives $p$ constrained: $(x-p)^2 - 2A p \ge 2Ax + A^2$ . Hmm. Let’s instead show $P$ bounded above directly: pick the $A$ -orbit $\{y_1+mA\}$ ; for $y$ large ( $y\ge y_1$ ), choose orbit point $x\in[y,y+A)$ ; if $d(y)=0$ , (L’‘): $(x-y)^2 \ge 2A(x+y)+A^2 > A^2$ , so $x - y > A$ , contradiction. Hence $d(y)=A$ for all $y \ge y_1$ . So $P \subseteq (0,y_1)$ , bounded ✓. - Let $s=\sup P \in (0, y_1]$ . Choose $p\in P$ with $p > s/2$ hmm need $p$ close to $s$ : choose $p\in P$ , $p > s - \epsilon$ . Choose $z \in Z$ (defect $A$ ) with $s < z < s+\epsilon$ — wait, is $(s, s+\epsilon)\subseteq Z$ ? Points $> s$ are not in $P$ (as $s=\sup P$ ), so they’re in $Z$ ✓ (using dichotomy: every point has defect $0$ or $A$ ). So $z\in Z$ with $z<s+\epsilon$ . - Now (R’’) with $x=p$ , $y=z$ : $(\sqrt p - \sqrt{z+A})^2 \ge d(z)-d(p) = A$ . But $z + A < s+\epsilon + A$ and $p > s-\epsilon$ , so $\sqrt{z+A}-\sqrt p < \sqrt{s+A+\epsilon} - \sqrt{s-\epsilon}$ . As $\epsilon\to0$ this tends to $\sqrt{s+A}-\sqrt s$ , and $(\sqrt{s+A}-\sqrt s)^2 = 2s + A - 2\sqrt{s(s+A)} < A$ since $\sqrt{s(s+A)} > s$ . So for small $\epsilon$ , $(\sqrt{z+A}-\sqrt p)^2 < A$ : contradiction ✓.

Good. So $d$ is constant: $d \equiv c \ge 0$ , $f(x)=x+c$ . And all such work.

Hmm wait, in Case 2, I assumed the value set is $\{0,A\}$ — justified: Suppose $d$ non-constant. If $d$ takes two distinct positive values, Case 1 gives contradiction. So $d$ takes at most one positive value $A$ , and since non-constant it takes value 0 too (as $d\ge0$ ), so value set $=\{0,A\}$ , $A>0$ . Then Case 2 gives contradiction. Hence $d$ constant ✓.

Wait, also in Case 1 I should double-check “two distinct positive values $A>B>0$ ” — yes, orbits of both go to infinity, argument works.

Hmm, but hold on. Let me double check whether Case 2 can be simplified: maybe we can avoid the sup argument with a cleverer use. Alternative for Case 2: We showed $d(y)=A$ for all $y\ge y_1$ (where $d(y_1)=A$ ). Hmm, actually more strongly, from (C): for all $y \ge y_1$ , $A - d(y) < \frac{A^2}{4y}$ , so with dichotomy $d(y)=A$ . Now let $p\in P$ , so $p<y_1$ . Apply (R’’) with $x=p$ and $y$ = any point $\ge y_1$ : $(\sqrt p - \sqrt{y+A})^2 \ge A$ : satisfied for large $y$ . So need points with defect $A$ near $p$ : that’s what the sup argument provides. OK keep sup argument.

Alternatively: Is there a slicker global argument? Let’s reconsider: maybe prove directly $f(x)-x$ is constant using (L’‘) and (R’’) with clever substitutions, avoiding case analysis.

Try: For any $x,y$ , (R’‘): $d(y) - d(x) \le (\sqrt x - \sqrt{f(y)})^2$ . Put $x = f(y) + h$ … then $d(y)-d(f(y)+h) \le (\sqrt{f(y)+h}-\sqrt{f(y)})^2 = \frac{h^2}{(\sqrt{f(y)+h}+\sqrt{f(y)})^2}\le \frac{h^2}{4f(y)}$ for $h> -f(y)$ . Hmm and (L’’) with $x = f(y)+h$ : $h^2 \ge 2(d(x)-d(y))(2f(y)+h) + (d(x)-d(y))^2$ , so if $d(x)>d(y)$ : $d(x)-d(y)\le \frac{h^2}{2(2f(y)+h)}\approx \frac{h^2}{4f(y)}$ . Combining: $|d(f(y)+h) - d(y)| \le \frac{h^2}{4f(y)}$ -ish, i.e., $d$ is “locally almost constant” near $f(y)$ , with modulus $h^2/(4f(y))$ .

Precisely: For all $y>0$ and $h$ with $f(y)+h>0$ , let $x=f(y)+h$ , $\delta = d(x)-d(y)$ . - If $\delta>0$ : (L’‘): $h^2 \ge 2\delta(x+f(y)) + \delta^2 \ge 2\delta \cdot (2f(y)+h)$ hmm $x+f(y) = 2f(y)+h$ . So $\delta \le \frac{h^2}{2(2f(y)+h)}$ . - If $\delta<0$ : (R’’): $-\delta \le (\sqrt x - \sqrt{f(y)})^2 = \frac{h^2}{(\sqrt{f(y)+h}+\sqrt{f(y)})^2}$ .

So in both cases $|\delta| \le \frac{h^2}{c\cdot f(y)}$ for appropriate constant when $|h|\le f(y)$ say. Then: $d$ is locally Lipschitz-ish, in fact differentiable with derivative 0 → constant! Because: the bound $|d(u)-d(v)|\le \frac{(u-v)^2}{C\min(u,v)}$ for nearby $u,v$ ? Careful: the bound compares $d$ at $x$ with $d$ at $y$ , where $x$ near $f(y) = y + d(y)$ , not near $y$ . Hmm. So it says: $d(f(y)+h) \approx d(y)$ for small $h$ . Since $d(f(y)) = d(y)$ (orbit), it says $d$ is almost constant near the point $f(y)$ , i.e. near any point in the image of $f$ . Then to conclude $d$ constant globally we need image of $f$ to be everything-ish. Hmm, image of $f$ = $\{y+d(y)\}$ . Not obviously all of $(0,\infty)$ .

But: it gives: for $u$ in the image of $f$ , $|d(u+h)-d(u)|\le \frac{h^2}{2u}$ roughly (for $|h|\le u$ ). Hmm, but to chain we need $u+h$ also in image… Suppose $u = f(y)$ ; then $u \in$ image, and $d(u)=d(y)$ ; also $f(u) = u + d(u) \in$ image. Hmm.

I think the case analysis proof is fine and rigorous. But let me reconsider — maybe there’s an even cleaner ending. Let’s think about the structure: we have $d\ge0$ , and

$\forall y, \forall$ small $h$ : $|d(f(y)+h) - d(y)|$ small.

Let $c = \inf d \ge 0$ . Take $y$ with $d(y)$ close to $c$ …

Alternatively: prove $d$ is constant via: for any $x,y$ , apply (L’‘)/(R’’) to pairs $(x, y)$ where we choose $y$ in $x$ ’s vicinity… The trouble is asymmetry: constraints involve $x$ vs $f(y)$ .

Given the problem likely from a competition (maybe the answer is $f(x)=x$ ?). Let me double check the problem: maybe the intended answer is indeed $f(x)=x$ and I mis-simplified. Let’s recheck $f(x) = x+1$ in the original with specific numbers: $x=1,y=1$ : $f(1)=2$ , $f(y)=2$ . Left: $\sqrt{(1+4)/2}=\sqrt{2.5}\approx1.581$ . Middle: $(2+1)/2=1.5$ . Right: $\sqrt{1\cdot2}=1.414$ . Indeed $1.581\ge1.5\ge1.414$ ✓.

Another: $x=4,y=1$ : $f(4)=5$ , $f(1)=2$ . Left: $\sqrt{(16+4)/2}=\sqrt{10}=3.162$ . Middle: $(5+1)/2=3$ . Right: $\sqrt{4\cdot2}=2.828$ ✓.

$x=0.1, y=5$ : $f(0.1)=1.1$ , $f(5)=6$ . Left: $\sqrt{(0.01+36)/2}=\sqrt{18.005}=4.243$ . Middle: $(1.1+5)/2=3.05$ . Right: $\sqrt{0.1\cdot 6}=0.7746$ ✓.

Yes so $f(x)=x+c$ works for any $c\ge 0$ . So the answer: $f(x) = x + c$ , $c\ge 0$ . Hmm, but is this a known problem? Possibly this is a variant. Actually I recall Iran TST or similar with answer $f(x)=x$ . Let me reconsider whether the problem might mean the inequalities hold for all $x,y$ and the answer is $f(x)=x+c$ … Since our derivation is solid, answer: $f(x)=x+c$ , $c\ge0$ (equivalently all functions $f(x)=x+c$ with $c \ge 0$ ).

Hmm wait, let me double check the step deriving $f(f(y)) = 2f(y)-y$ : Set $x = f(y)$ in the given chain. Then LHS $=\sqrt{\frac{f(y)^2+f(y)^2}{2}} = f(y)$ . RHS $=\sqrt{f(y)f(y)} = f(y)$ . Middle $=\frac{f(f(y))+y}{2}$ . So $f(y) \ge \frac{f(f(y))+y}{2} \ge f(y)$ , forcing equality: $f(f(y)) + y = 2f(y)$ ✓✓.

Then orbit AP and positivity ⟹ $d\ge0$ ✓.

Now let me also double check (L’‘) and (R’’) derivations. (L): $\frac{x^2+f(y)^2}{2} \ge \frac{(f(x)+y)^2}{4}$ ⟺ $2x^2+2f(y)^2 \ge (f(x)+y)^2$ . With $u=x$ , $v=f(y)$ : $f(x)+y = x + d(x) + y = u + (v - d(y)) + d(x) = u+v+\delta$ where $\delta = d(x)-d(y)$ ✓ (since $y = f(y)-d(y) = v-d(y)$ ). So (L) ⟺ $2u^2+2v^2 \ge (u+v+\delta)^2$ ⟺ $(u-v)^2 \ge 2\delta(u+v)+\delta^2$ ✓. (R): $\frac{f(x)+y}{2}\ge\sqrt{xf(y)}$ ⟺ $u+v+\delta \ge 2\sqrt{uv}$ ⟺ $(\sqrt u-\sqrt v)^2 \ge -\delta$ ✓.

Both hold for all $x,y>0$ , where $u=x$ ranges over all positive reals, $v = f(y)$ ranges over $f(\mathbb R_{>0})$ , $\delta = d(x)-d(y)$ .

Nice, so combined: for all $x,y>0$ , with $\delta = d(x)-d(y)$ , $u=x$ , $v=f(y)$ : $(u-v)^2 \ge 2\delta(u+v)+\delta^2 \quad\text{and}\quad (\sqrt u - \sqrt v)^2 \ge -\delta.$

Note the second can be rewritten $(u-v)^2 \ge -\delta(\sqrt u+\sqrt v)^2$ , hmm. Combined statement: $(u-v)^2 \ge$ … Let’s unify: For $\delta\ge0$ : $(u-v)^2 \ge 2\delta(u+v)$ . For $\delta\le 0$ , with $\epsilon=-\delta\ge0$ : $(\sqrt u-\sqrt v)^2\ge \epsilon$ so $(u-v)^2 = (\sqrt u-\sqrt v)^2(\sqrt u+\sqrt v)^2 \ge \epsilon(\sqrt u+\sqrt v)^2\ge \epsilon(u+v)$ hmm since $(\sqrt u+\sqrt v)^2 = u+v+2\sqrt{uv}\ge u+v$ . So in both cases: $(x - f(y))^2 \ \ge\ |d(x)-d(y)|\,\big(x+f(y)\big) \quad \forall x,y>0. \tag{★}$ Nice single clean inequality! (For $\delta\ge0$ we even have factor 2.)

Now with (★) the endgame: Suppose $d(y_1)=A > B=d(y_2)$ , so $A>0$ .

Using orbits: $d(y_1+mA)=A$ , $f(y_1+mA) = y_1+(m+1)A$ ; similarly for $y_2$ if $B>0$ .

Endgame with (★): Case $B>0$ : take $y = y_2+kB$ , then $f(y) = y_2+(k+1)B =: V_k\to\infty$ ; choose $x = y_1+mA$ with $|x - V_k| \le A$ (possible for $V_k \ge y_1$ ). Then (★): $A^2 \ge (A-B)(x + V_k) \ge (A-B)V_k \to\infty$ : contradiction ✓.

Case $B = 0$ : (so $y_2$ is a fixed point, $f(y_2)=y_2$ ). Sub-step 1: For all $y\ge y_1$ : $d(y) > A - \frac{A^2}{4y}$ ? Let’s derive with (★): choose $x=y_1+mA$ with $x \in [f(y), f(y)+A)$ , hmm we need $x$ near $f(y)$ : (★) with this $x$ : $A^2 > (x-f(y))^2 \ge |A - d(y)|(x+f(y)) \ge |A-d(y)|\cdot 2f(y) \ge |A-d(y)|\cdot 2y$ . So $|A - d(y)| < \frac{A^2}{2y}$ for all $y$ with $f(y)\ge y_1$ , in particular all $y \ge y_1$ . So $d(y)\to A$ as $y\to\infty$ ; more precisely $|d(y)-A| < \frac{A^2}{2y}$ for $y\ge y_1$ . Sub-step 2: fixed point $y_2$ : apply (★) with $x = y_2$ , $y$ arbitrary: $(y_2 - f(y))^2 \ge |d(y_2)-d(y)|(y_2+f(y)) = d(y)(y_2+f(y)) \ge d(y) f(y)$ . Hmm, choose $y$ with $f(y)$ close to $y_2$ ! Then LHS small, and $d(y)\ge$ ? Need $d(y)$ bounded below for such $y$ . Hmm.

Alternatively Sub-step 2’: Use (★) with $x$ near $y_2$ ?? The pair $(x,y)$ constraint involves $x$ vs $f(y)$ . To exploit the fixed point we can set $y = y_2$ (giving $f(y)=y_2$ ) and $x$ arbitrary: $(x-y_2)^2 \ge d(x)(x+y_2)$ . So $d(x) \le \frac{(x-y_2)^2}{x+y_2}\quad\forall x. \tag{†}$ Combined with sub-step 1: for $x\ge y_1$ : $d(x) > A - \frac{A^2}{2x}$ . Is there a conflict? (†) at large $x$ gives $d(x)\lesssim x$ : no conflict. Need conflict at moderate $x$ : we need $x$ where $\frac{(x-y_2)^2}{x+y_2} < A - \frac{A^2}{2x}$ , and $x \ge y_1$ . Since $x\ge y_1 > y_2$ ? Not necessarily: is $y_1 > y_2$ ? Unknown. Hmm. If we could apply sub-step 1’s conclusion for $x$ near $y_2$ we’d be done: at $x=y_2$ : (†) gives $d(y_2)=0$ consistent. For $x$ slightly $> y_2$ , (†) gives $d(x)$ tiny, but sub-step 1 needs $x\ge y_1$ ; if $y_1 \le y_2$ then… but note: from sub-step 1 conclusion $|d(y)-A|<A^2/(2y)$ for $y\ge y_1$ ; taking $y=y_2$ (if $y_2\ge y_1$ ): $|0 - A| < A^2/(2y_2)$ ⟹ $y_2 < A/2$ . Hmm, so a fixed point must be $< A/2$ and $<y_1$ .

So: fixed points are all $< A/2$ and $< y_1$ ? Wait we need $y_2 \ge y_1$ for that; so we get: every fixed point $p$ satisfies $p< y_1$ or $p<A/2$ . Hmm, but actually we can improve: use sub-step 1 with better orbit base. Note $d(y)>A - A^2/(2y)>0$ for $y > A/2$ , $y\ge y_1$ . So all fixed points are in $(0,\max(y_1, A/2))$ . Hmm, that’s not $(0,A/2)$ unless $y_1$ small.

But: replace $y_1$ by a bigger orbit? We can also use, instead of the orbit of $y_1$ , other points with defect close to $A$ … Let’s think again: from sub-step 1, $d(y)\to A$ as $y\to\infty$ . Consider $y_1’ = $ some large point; $A' = d(y_1')$ close to $A$ . Its orbit gives $|d(y)-A'|<\frac{A'^2}{2y}$ for $y \ge y_1'$ . Doesn’t help push down.

OK here’s the thing: to handle case $B=0$ we used earlier the dichotomy of values (only $\{0,A\}$ ) and the sup argument with (R’’). Let’s redo that with (★) and the general setting (not needing dichotomy?).

Set $P = \{p>0: d(p)=0\}$ ; assume $P\ne\emptyset$ and $d\not\equiv0$ . From (†) applied to any $p \in P$ : $d(x)\le \frac{(x-p)^2}{x+p}$ for all $x$ — this holds for every $p\in P$ ! Let $s=\sup P$ . From sub-step 1 with any positive-defect point (exists), $P$ is bounded: indeed $d(y)>0$ for $y \ge \max(y_1, A/2)$ , so $s \le \max(y_1,A/2)<\infty$ .

Now take $p\in P$ with $p$ close to $s$ , and $x$ slightly $>s$ hmm need lower bound on $d(x)$ for $x$ slightly above $s$ . But points just above $s$ have positive defect (not in $P$ ) but possibly tiny defect — no contradiction! Hmm! So the dichotomy was essential. But without dichotomy, could there be a function with defects tapering to 0? Let’s reconsider: maybe the general argument needs more care. Let’s think about what (★) plus orbit structure really forces.

Let’s reconsider. Facts: 1. $d\ge 0$ , $d(y+kd(y)) = d(y)$ for all $k\in\mathbb Z_{\ge0}$ . 2. (★): $(x-f(y))^2 \ge |d(x)-d(y)|(x+f(y))$ .

Suppose $d$ non-constant. Let $A = \sup d$ , possibly $\infty$ ? Let’s see: (†)-like bound from any point: (★) with general $y$ : $d(x) \le d(y) + \frac{(x-f(y))^2}{x+f(y)}$ . Fix $y=y_0$ : $d(x) \le d(y_0) + \frac{(x - f(y_0))^2}{x+f(y_0)} \le d(y_0)+x$ roughly. So $d$ has at most linear growth; sup could be infinite.

Let’s establish: (Sub-step 1 generalized) For any $y_1$ with $d(y_1)=A_1>0$ , and any $y$ with $f(y)\ge y_1$ : $|d(y)-A_1| \le \frac{A_1^2}{2f(y)}$ hmm let me redo: choose $x$ in orbit of $y_1$ with $x - f(y)\in[0,A_1)$ (possible when $f(y)\ge y_1$ ). Then $(x-f(y))^2 < A_1^2$ and (★): $A_1^2 > |d(x)-d(y)|(x+f(y)) \ge |A_1-d(y)|\cdot 2f(y)$ . So $|d(y) - A_1| < \frac{A_1^2}{2f(y)} \le \frac{A_1^2}{2y}\quad \text{whenever } f(y)\ge y_1. \tag{C'}$ In particular for all $y\ge y_1$ (since $f(y)\ge y$ ).

Now: this says any two positive-defect values $A_1, A_2$ (at $y_1,y_2$ ) satisfy: for large $y$ , $d(y)$ is within $A_1^2/(2y)$ of $A_1$ and within $A_2^2/(2y)$ of $A_2$ ; let $y\to\infty$ : $A_1 = A_2$ ! So all positive defect values are equal ✓ (this handles case 1 cleanly, and gives dichotomy: $d$ takes values in $\{0,A\}$ where $A>0$ is the unique positive value).

Then the sup argument with (R’‘) finishes. Good — so dichotomy comes free from (C’), and then $P$ bounded, $s=\sup P$ , points above $s$ have defect exactly $A$ (dichotomy!), and (R’’) contradiction.

Let me now write the (R’‘) final contradiction carefully. $s = \sup P \in (0,\infty)$ . For any $\epsilon>0$ : pick $p \in P$ with $p > s-\epsilon$ . Pick $z = s + \epsilon$ hmm need $z\notin P$ : since $z > s$ , $z\notin P$ , so $d(z) = A$ . Apply (R’’): $(\sqrt x - \sqrt{f(y)})^2 \ge d(y) - d(x)$ with $x = p$ , $y=z$ : $d(y)-d(x) = A - 0 = A$ , $f(z) = z+A = s+\epsilon+A$ . So need $\left(\sqrt{s+\epsilon+A} - \sqrt p\right)^2 \ge A.$ Since $p > s-\epsilon$ : LHS $< (\sqrt{s+A+\epsilon}-\sqrt{s-\epsilon})^2$ (assuming $s-\epsilon>0$ and $\sqrt{s+A+\epsilon}>\sqrt{p}$ ✓). As $\epsilon\to 0^+$ , this tends to $(\sqrt{s+A}-\sqrt s)^2 = 2s+A - 2\sqrt{s(s+A)} < 2s + A - 2s = A$ . So for small enough $\epsilon$ , we get LHS $< A$ , contradicting the requirement. Hence $P=\emptyset$ or $d\equiv0$ …

Precisely: contradiction shows: cannot have both $P\ne\emptyset$ and $d$ taking positive value $A$ . So either $d\equiv 0$ ( $f(x)=x$ ) or $d$ never $0$ hence $d\equiv A$ constant.

Wait: if $P=\emptyset$ then $d$ takes only positive values, all equal by (C’) ⟹ $d\equiv A$ ✓. If $P \ne\emptyset$ and some positive value exists → contradiction. If $P\ne\emptyset$ and no positive values → $d\equiv0$ ✓.

So $f(x) = x+c$ , $c\ge0$ . And verification shows all such work.

Hmm, let me double-check the claim $(\sqrt{s+A}-\sqrt s)^2 < A$ : expand $= s + A - 2\sqrt{s(s+A)} + s = 2s+A-2\sqrt{s^2+sA}$ . Since $\sqrt{s^2+sA} > \sqrt{s^2} = s$ (as $sA>0$ ), we get $<2s+A-2s = A$ ✓.

Also need continuity in $\epsilon$ : define $g(\epsilon) = (\sqrt{s+A+\epsilon}-\sqrt{s-\epsilon})^2$ , continuous, $g(0)<A$ , so exists $\epsilon_0>0$ with $g(\epsilon)<A$ for $\epsilon<\epsilon_0$ ✓. Also need $s-\epsilon>0$ ✓ for small $\epsilon$ .

One more check: In (C’) derivation I need orbit of $y_1$ to reach beyond $f(y)$ : the orbit is $y_1 + mA_1$ , $m\ge 0$ , with $d = A_1$ (proved by induction from $f(f(y))=2f(y)-y$ ). Given target $T=f(y) \ge y_1$ , choose $m = \lfloor (T-y_1)/A_1\rfloor$ , then $x=y_1+mA_1 \le T < x + A_1$ , so $T - x \in [0,A_1)$ ✓. (I earlier wrote $x-f(y)\in[0,A_1)$ ; either way $|x-T|<A_1$ ✓.)

Then (★) with this $x$ and $y$ : $(x-f(y))^2 \ge |d(x)-d(y)|(x+f(y))$ ; $d(x)=A_1$ ; $x + f(y) \ge f(y) + f(y) - A_1$ hmm: $x > T - A_1 = f(y)-A_1$ , so $x+f(y) > 2f(y)-A_1$ . Let’s be safe: $x + f(y) \ge f(y)$ (since $x>0$ ). Then $|A_1 - d(y)| \le \frac{(x-f(y))^2}{x+f(y)} < \frac{A_1^2}{f(y)}$ . Good enough: $|d(y)-A_1| < \frac{A_1^2}{f(y)} \le \frac{A_1^2}{y}, \quad \text{for } f(y)\ge y_1.$ Then two positive values: for $y\to\infty$ , $|d(y)-A_1|\to0$ and $|d(y)-A_2|\to 0$ ⟹ $A_1=A_2$ ✓. (Just take a sequence $y\to\infty$ , e.g. $y = n$ .)

Also need $P$ bounded: from above, for $y \ge \max(y_1, 2A_1^2/A_1)$ hmm: $d(y) > A_1 - A_1^2/y > 0$ when $y > A_1$ . So for $y\ge\max(y_1, A_1)$ … take $y > \max(y_1,A_1)$ : $d(y)>0$ , so $P \subseteq (0, \max(y_1,A_1)]$ ✓ bounded, and $s=\sup P$ finite and positive (as $P\ne\emptyset$ contains a positive number).

Everything checks. Also should double check that (★) holds for all $x,y$ — derived from (L) and (R) which hold for all $x,y$ ✓, using $d(x)=f(x)-x$ , $\delta = d(x)-d(y)$ , and note $y = f(y)-d(y)$ ✓.

Also, I should double check the claim $d\ge0$ carefully: from $f(f(y))=2f(y)-y$ , define $y_0=y$ , $y_{n+1}=f(y_n)$ . Then $y_{n+2}=f(f(y_n))=2f(y_n)-y_n = 2y_{n+1}-y_n$ . So $y_{n+1}-y_n$ constant $= y_1-y_0 = d(y)$ . Thus $y_n = y + n d(y) > 0$ for all $n$ (since $f$ maps into $\mathbb R_{>0}$ ). If $d(y)<0$ , then $y_n<0$ for large $n$ : contradiction ✓. So $d(y)\ge0$ and $f(y+nd(y)) = y+(n+1)d(y)$ , i.e. $d(y+nd(y))=d(y)$ ✓.

Now, write up the solution.

Write-up plan:

Answer: $f(x)=x+c$ for some constant $c\ge0$ .

Verification: For such $f$ , with $a=x>0$ , $b=f(y)=y+c>0$ : the middle term is $\frac{f(x)+y}{2}=\frac{x+c+y}{2}=\frac{a+b}{2}$ , left is $\sqrt{\frac{a^2+b^2}{2}}$ , right is $\sqrt{ab}$ ; chain is QM ≥ AM ≥ GM ✓.

Necessity: Step 0: Notation: write (L), (R). Step 1: $x=f(y)$ ⟹ $f(f(y))=2f(y)-y$ . Step 2: orbits are APs ⟹ $d:=f-\mathrm{id}\ge0$ and $d(y+nd(y))=d(y)$ , $f(y+nd(y))=y+(n+1)d(y)$ . Step 3: reformulate (L),(R) as $(u-v)^2\ge 2\delta(u+v)+\delta^2$ and $(\sqrt u-\sqrt v)^2\ge-\delta$ with $u=x$ , $v=f(y)$ , $\delta=d(x)-d(y)$ ; deduce (★): $(x-f(y))^2 \ge |d(x)-d(y)|(x+f(y))$ . Step 4: (C’): if $d(y_1)=A>0$ then $|d(y)-A|<A^2/f(y)\le A^2/y$ for all $y$ with $f(y)\ge y_1$ . Step 5: All positive values of $d$ coincide; $P=d^{-1}(0)$ bounded. Step 6: If $P\ne\emptyset$ and $A>0$ exists: contradiction via (R) at $x=p\approx s=\sup P$ , $y=z$ slightly above $s$ . Step 7: Conclude $d\equiv c$ .

Let me double-check Step 6’s use of (R’‘): (R’’) is $(\sqrt x - \sqrt{f(y)})^2 \ge d(y)-d(x)$ . Direct from (R): $\frac{f(x)+y}{2}\ge\sqrt{xf(y)}$ ⟺ $f(x)+y\ge 2\sqrt{x f(y)}$ ⟺ $x + d(x) + f(y)-d(y) \ge 2\sqrt{x f(y)}$ ⟺ $(\sqrt x - \sqrt{f(y)})^2 \ge d(y)-d(x)$ ✓.

With $x=p$ ( $d(p)=0$ ), $y=z$ ( $d(z)=A$ , $f(z)=z+A$ ): $(\sqrt p - \sqrt{z+A})^2 \ge A$ ✓ as used.

Also, in Step 6, I need $d(z)=A$ for $z>s$ : since $z\notin P$ means $d(z)>0$ , and by Step 5 all positive values equal $A$ ✓.

Also need in step 5: existence of at least one positive value if $d\not\equiv0$ ✓.

Let me also double check Step 4 more carefully including edge cases: $y_1$ with $A=d(y_1)>0$ . Orbit points $x_m=y_1+mA$ have $d(x_m)=A$ and are $\ge y_1$ , spacing $A$ , unbounded. For $y$ with $T:=f(y)\ge y_1$ : choose $m=\lfloor (T-y_1)/A \rfloor \ge 0$ : $x_m \le T < x_m + A$ so $0\le T-x_m<A$ . (★) with $x=x_m$ : $(x_m - T)^2 \ge |A - d(y)|(x_m+T)$ . Since $(x_m-T)^2 < A^2$ and $x_m + T > T$ : $|A-d(y)| < A^2/T$ ✓.

Good. All solid. Note also Step 4 gives another proof that $P$ bounded and dichotomy.

Now also double-check the final answer includes $c=0$ i.e. $f(x)=x$ ✓.

I’ll also remark that (★) with factor 2 for $\delta\ge0$ isn’t needed.

Let me write the final solution nicely.

Also perhaps present the alternate: the problem might expect answer $f(x)=x+c$ . I’ll present full rigor.

Write-up details for Step 3 (★): Let $u=x>0$ , $v=f(y)>0$ , $\delta=d(x)-d(y)$ . Since $y=v-d(y)$ and $f(x)=u+d(x)$ : $f(x)+y = u+v+\delta$ . (L) ⟺ $2(u^2+v^2)\ge (u+v+\delta)^2$ . Using $2(u^2+v^2)=(u+v)^2+(u-v)^2$ and $(u+v+\delta)^2 = (u+v)^2+2\delta(u+v)+\delta^2$ : ⟺ $(u-v)^2\ge 2\delta(u+v)+\delta^2$ . (R) ⟺ $u+v+\delta \ge 2\sqrt{uv}$ ⟺ $(\sqrt u-\sqrt v)^2 \ge -\delta$ . If $\delta\ge0$ : $(u-v)^2 \ge 2\delta(u+v) \ge \delta(u+v)$ . If $\delta<0$ : $(u-v)^2 = (\sqrt u-\sqrt v)^2(\sqrt u+\sqrt v)^2 \ge (-\delta)(u+v+2\sqrt{uv}) \ge |\delta|(u+v)$ . Hence (★) ✓.

Ready to write final answer.