2026 IMO Problem 5 — Natural-Language Solution

Problem

Determine all functions

f : \mathbb{R}_{>0}\to\mathbb{R}_{>0}

such that, for all $x,y>0$ ,

\sqrt{\frac{x^2+f(y)^2}{2}} \ge \frac{f(x)+y}{2} \ge \sqrt{x f(y)}.

Solution

All solutions are

\boxed{f(x)=x+c\quad(c\ge0)}.

For $x>0$ , define the displacement

d(x)=f(x)-x.

A quantitative displacement estimate

Squaring the two given inequalities is legitimate because all quantities involved are nonnegative. Put $S=f(x)+y$ and $T=x+f(y)$ . The squared inequalities give

4x f(y)\le S^2\le 2\bigl(x^2+f(y)^2\bigr),

and hence

-(x-f(y))^2\le S^2-T^2\le(x-f(y))^2.

Since

S^2-T^2 =\bigl(d(x)-d(y)\bigr)\bigl(f(x)+y+x+f(y)\bigr),

and the second factor is positive, taking absolute values gives the main estimate, labeled $(1)$ :

|d(x)-d(y)|\bigl(f(x)+y+x+f(y)\bigr) \le (x-f(y))^2.

Set $x=f(y)$ in the original inequalities. The two outer square roots both become $f(y)$ , so equality is forced in the middle and

f(f(y))-f(y)=f(y)-y.

Consequently every forward orbit is an arithmetic progression; we label this identity $(2)$ :

f^{\,n}(x)=x+n\,d(x).

Here $f^{\,n}$ denotes the $n$ -fold iterate of $f$ , with $f^{\,0}$ equal to the identity.

Since every iterate remains positive, $(2)$ rules out $d(x)<0$ ; otherwise the right-hand side would eventually be negative. Hence we have $(3)$ :

d(x)\ge0\qquad\text{for every }x>0.

The displacement is constant

First suppose $d(x)=a>0$ and $d(y)=b>0$ . Formula $(2)$ gives two positive arithmetic orbits with steps $a$ and $b$ ; in particular, the displacement is $a$ at every point of the first orbit and $b$ at every point of the second. If $a\ne b$ , choose a sufficiently distant point $Y$ on the $b$ -orbit, put $V=f(Y)=Y+b$ , and then choose the last point $U$ of the $a$ -orbit not exceeding $V$ . Thus

U\le V<U+a.

The distance $V-U$ is less than $a$ , while $V$ can be made so large that

a^2<|a-b|V.

Applying $(1)$ to $U$ and $Y$ gives

|a-b|V <|a-b|\bigl(f(U)+Y+U+V\bigr) \le(U-V)^2<a^2,

a contradiction. Therefore all positive values of $d$ are equal.

Let their common positive value, if one exists, be $a$ . If $d(p)=a$ and $d(q)=0$ , estimate $(1)$ gives the separation bound $(4)$ :

|p-q|\ge a.

Indeed, apply $(1)$ with $x=p$ and $y=q$ . Since $f(p)=p+a$ and $f(q)=q$ , the positive factor multiplying $a$ on the left is greater than $a$ . Thus, if $|p-q|<a$ , the left side of $(1)$ is strictly greater than $a^2$ , while its right side is strictly less than $a^2$ .

Thus $d$ takes only the values $0$ and $a$ . By $(4)$ , no two points less than $a$ apart can carry different values, so $d$ is locally constant on the connected interval $(0,\infty)$ . A locally constant function on a connected interval is constant. If no positive displacement exists, $(3)$ already implies $d\equiv0$ . In all cases there is a constant $c\ge0$ such that

d(x)=c, \qquad\text{so}\qquad f(x)=x+c.

Verification

Conversely, let $f(x)=x+c$ with $c\ge0$ . The two required inequalities become

\sqrt{\frac{x^2+(y+c)^2}{2}} \ge \frac{x+y+c}{2} \ge\sqrt{x(y+c)}.

The first is the quadratic-mean/arithmetic-mean inequality, and the second is the arithmetic-mean/geometric-mean inequality, both applied to the nonnegative numbers $x$ and $y+c$ . Hence every such translation works.