IMO 2026 · Problem 5

Let \(\mathbb R_{>0}\) be the set of positive real numbers. Determine all functions \(f\colon\mathbb R_{>0}\to\mathbb R_{>0}\) such that \[ \sqrt{\frac{x^2+f(y)^2}{2}}\ \ge\ \frac{f(x)+y}{2}\ \ge\ \sqrt{x f(y)} \] for every \(x,y\in\mathbb R_{>0}\).

Official text of the problem. Official problem paper (PDF)

Answer

\(\boxed{f(x)=x+c\text{ for all }x>0,\text{ where }c\ge0\text{ is a constant}.}\)

Solution

Suppose \(f\) satisfies the given inequalities. Substituting \(x=f(t)\) and \(y=t\), both outer expressions equal \(f(t)\). Thus \[ f(f(t))+t=2f(t). \] In particular, positivity of \(f(f(t))\) gives \[ f(t)>\frac t2 \qquad(t>0). \tag{1} \]

Next, apply the right-hand inequality with \(x=f(a)\) and \(y=b\), where \(a,b>0\). Squaring and using the identity above yields \[ 4f(a)f(b) \le \bigl(f(f(a))+b\bigr)^2 =\bigl(2f(a)+b-a\bigr)^2. \] Dividing by \(4f(a)>0\) and rearranging, we obtain \[ \bigl(f(b)-b\bigr)-\bigl(f(a)-a\bigr) \le \frac{(b-a)^2}{4f(a)} \le \frac{(b-a)^2}{2a}, \] where the last inequality follows from (1).

Define \(g(t)=f(t)-t\). Interchanging \(a\) and \(b\) in the preceding bound shows that \[ |g(b)-g(a)| \le \frac{(b-a)^2}{2\min\{a,b\}}. \tag{2} \]

We now prove that \(g\) is constant. Fix \(0<a<b\). For any positive integer \(n\), partition \([a,b]\) into equal parts with endpoints \[ t_i=a+\frac{i(b-a)}n,\qquad i=0,1,\ldots,n. \] By (2) and the triangle inequality, \[ \begin{aligned} |g(b)-g(a)| &\le \sum_{i=1}^{n}|g(t_i)-g(t_{i-1})|\\ &\le \sum_{i=1}^{n}\frac{(t_i-t_{i-1})^2}{2a} =\frac{(b-a)^2}{2an}. \end{aligned} \] Letting \(n\to\infty\) gives \(g(b)=g(a)\). Hence \(f(x)=x+c\) for some real constant \(c\). Positivity for every \(x>0\) forces \(c\ge0\): if \(c<0\), taking \(x=-c/2\) would give \(f(x)=c/2<0\).

Conversely, let \(c\ge0\) and \(f(x)=x+c\). For \(u=x>0\) and \(v=y+c>0\), we have \(f(x)+y=u+v\) and \(f(y)=v\). The required inequalities become \[ \sqrt{\frac{u^2+v^2}{2}} \ge \frac{u+v}{2} \ge \sqrt{uv}. \] Both hold because \[ 2(u^2+v^2)-(u+v)^2=(u-v)^2\ge0, \qquad (u+v)^2-4uv=(u-v)^2\ge0. \] This verifies every function in the stated family.

The solution was checked against Evan Chen's published IMO solution notes.