IMO 2026 | General Regulations

Solution 1
We claim $f(x) = x+c$ for $c\geq 0$ works. This clearly does due to QM-AM-GM.

 

Now I'll prove its the only solution. Let $P(x,y)$ be the assertion. Subbing $P(f(x),x)$ forces
\[f(x) \geq \frac{f(f(x))+x}{2}\geq f(x)\]So $f(f(x))= 2 f(x)-x$. Now, this means $f(f(x))-f(x)=f(x)-x=c$ and so spamming $c$ (considering $f^n(x)$ which is an AP) gives $cn + x \geq 0$ for large $n$, this is only possible for $c\geq 0$.

Thus $g(x) = f(x) -x \geq 0$ always. Note that $g(x+g(x))=g(x)$ as
\[f(x+g(x))- x-g(x) = f(f(x)) - f(x) = f(x) - x= g(x)\]Let's use the RHS. $P(x,y)$ now becomes:
\[\frac{x+g(x)+y}{2} \geq \sqrt{x(y+g(y)}\]Expanding,
\[x^2 + g(x)^2 + y^2 + 2g(x)(x+y)+2xy \geq 4xy + 4xg(y)\]\[x^2+y^2-2xy +g(x)^2 + 2g(x)(y-x) \geq 4x(g(y)-g(x))\]The LHS is \[(x-y-g(x))^2\]Now lets say that $g$ takes two different nonzero values. So let $g(a) = u$ and $g(b) = v$, wlog $u < v$. Our earlier recurison gives us $g(a+nu ) = u$ for all $n$ and similarly $g(b+mv) = v$. $P(a+nu, b+mv)$ hence yields
\[(a-b+nu-mv - u)^2 \geq 4(a+nu)(v-u)\]Let $m = \left\lfloor \frac{a - b - u + nu}{v} \right\rfloor $ wishfully so the LHS is less than $v^2$ which is a constant. Take $n\to \infty$, and we're done. (noting $m > 0$ and is integer for large $n$)

Hence $g(x)$ takes only two distinct values, $0$ and $c > 0$. Now, the LHS of $P(x,y)$ such $f(x) = x+c$ and $f(y)=y$ gives
\[(x-y)^2 \geq 2c(x+y) +c^2 > c^2\]hence $|x-y| > c$. So the distance between points shifted by $c$ and identity is $c$. This means that for given $y$ identity, all points with $|x-y| \leq c$ are identity and we can extend this for all positive reals. the same applies to points $x$ with $f(x)=x+c$, and this is a contradiction as $c\neq 0$.

So all of $f(x)$ is either just $x$ or just $x+c$, for $c > 0$.

Solution 2
The answer is $f(x) = x + c$ for all $x \in \mathbb{R}_{ > 0}$, for some $c \ge 0$. We can see this works as the statement naturally leads to QM-AM-GM, with the extra observation that $f(x) + y = x + f(y)$. We will now prove that there are no other functions that work. Let $P(x, y)$ be the assertion of $x$ and $y$ to the given functional equation.

 

The crux of the problem is to work with $g(x) := f(x) - x$ for all $x \in \mathbb{R}_{ > 0}$.

Claim 01. $g(x) \ge 0$ for all $x \in \mathbb{R}_{ > 0}$.
Proof. Indeed, $P(f(x), x)$ gives us $f(f(x)) = 2f(x) - x$. One can thus prove by induction that $f^{n}(x) = x + n(f(x) - x)$ for any $n \ge 1$. However, by taking $n \to \infty$, as LHS is positive, we thus conclude that $f(x) - x \ge 0$, i.e., $g(x) \ge 0$.

Claim 02. There exists at most one positive real number $\alpha$ for which $g(x) = \alpha$ for any $x \in \mathbb{R}_{ > 0}$.
Proof. Let us suppose otherwise, that there exists $x, y, \alpha, \beta \in \mathbb{R}_{ > 0}$ for which $g(x) = \alpha$ and $g(y) = \beta$, while $\alpha \not= \beta$. For all sufficiently large $m$, we define $\ell_m \ge 1$ to be such that
\[ \left| \ell_m - \frac{x - y + m \alpha}{\beta} \right| \le \frac{1}{2} \]and define $a_m := f^m(x), b_m := f^{\ell_m - 1}(y), c_m := f^{\ell_m}(y)$. Now, note that $P(a_m, b_m)$ gives us
\[ \sqrt{\frac{a_m^2 + c_m^2}{2}} \ge \frac{f(a_m) + b_m}{2} \ge \sqrt{a_m c_m}. \]We have seen above that $f^n(x) = x + n(f(x) - x)$ for any $n \ge 1$, which allows us to write $f(a_m) = a_m + \alpha$ and $b_m = c_m - \beta$. Therefore, this above inequality rewrites to
\[ \sqrt{\frac{a_m^2 + c_m^2}{2}} \ge \frac{a_m + c_m + (\alpha - \beta)}{2} \ge \sqrt{a_m c_m}. \]Now, it's time to do analysis: we basically can rewrite this as
\[ - \frac{(a_m - c_m)^2}{(\sqrt{a_m} + \sqrt{c_m})^2} = 2\sqrt{a_m c_m} - (a_m + c_m) \le \alpha - \beta \le \sqrt{2(a_m^2 + c_m^2)} -(a_m + c_m) = \frac{(a_m - c_m)^2}{\sqrt{2(a_m^2 + c_m^2)} + a_m + c_m}. \]However, by our construction,
\[ |a_m - c_m| = |f^m(x) - f^{\ell_m}(y)| = |x + m \alpha - y - \ell_m \beta| \le \frac{1}{2} \beta, \]which is bounded; and $a_m, c_m \to \infty$ since $\alpha, \beta > 0$. Therefore, by taking $m \to \infty$; both LHS and RHS goes to $0$. This shows $\alpha = \beta$, which is what we wanted to show. Claim 03. There is at most one possible value for $g(x)$ for any $x \in \mathbb{R}_{ > 0}$.
Proof. Suppose otherwise, then from the previous claim, we know that there exists $\alpha$ for which both the sets
\[ A_1 := \{ x \in \mathbb{R}_{ > 0} : g(x) = \alpha \} \ \text{and} \ A_2 := \{ x \in \mathbb{R}_{ > 0} : g(x) = 0 \} \]are both disjoint and nonempty. Furthermore, $P(x,y)$ whenever $x \in A_2$ and $y \in A_1$ gives us
\[ \alpha \le \sqrt{2(x^2 + y^2)} - (x + y). \]Now, we know that there exists $x_0 \in A_1$ and $y_0 \in A_2$ as both are nonempty. Now, we claim that we can always find $(x_n, y_n)$ with arbitrarily small distance where both of them lie on different set, while both of them stay bounded from below. Indeed, suppose $(x_n, y_n)$ have been chosen, then consider $\frac{x_n + y_n}{2}$ which must lie on either $A_1$ or $A_2$, and use this to replace $x_n$ or $y_n$ such that the condition still holds true. Note that this way, $|x_n - y_n| \to 0$ as $n \to \infty$. Furthermore, we know that $x_n, y_n \ge \min \{ x_0, y_0 \} > 0$ by construction and thus by plugging $(x_n, y_n)$ to the bound we have above, we get
\[ \alpha \le \frac{(x_n - y_n)^2}{\sqrt{2(x_n + y_n)^2} + x_n + y_n} \to 0 \]as $n \to \infty$, but $\alpha > 0$, which is a contradiction.

Solution 3
Answer is $f(x) \equiv x+c$, for any $c > 0$. This works, because \[ \sqrt{\frac{x^2 + (y+c)^2}{2}} \ge \underbrace{\frac{(x+c)+y}{2}}_{=\frac{x+(y+c)}{2}} \ge \sqrt{x \cdot (y+c)} \]is immediate by QM-AM-GM on $x$ and $y+c$.
For the converse direction, we let $P(x,y)$ denote the left inequality and $Q(x,y)$ denote the right inequality. We show two solutions. Iterated solution. By $f^n$ we mean $f$ applied $n$ times.
Claim: For any $t > 0$, the sequence \[ \mathbf{S}(t) = (t, f(t), f^2(t), f^3(t), \dots) \]forms an arithmetic progression (with nonnegative common difference).
Proof. Note that $P(f(t), t)$ and $Q(f(t), t)$ together gives the equality \[ \frac{f(f(t))+t}{2} = f(t) \]showing the first three terms are an arithmetic progression. More generally, applying $P(f^n(t), f^{n-1}(t))$ and $Q(f^n(t), f^{n-1}(t))$ for all $n \ge 2$ gives the full result. Since all terms of the sequence are positive, the common difference is nonnegative. $\blacksquare$
To finish the problem we just need to show all the common differences that arise this way are the same. The following result proves this except in an annoying edge case of fixed points (which we deal with in the final claim).
Claim: Choose any $x, y > 0$, and let $A$ and $B$ be the common differences of $\mathbf{S}(x)$ and $\mathbf{S}(y)$. If $A > 0$ and $B > 0$, then $A=B$.
Proof. We just prove $A \ge B$ (by symmetry); we only use $Q$ for this. For any positive integers $m$ and $n$, $Q(f^m(x), f^n(y))$ gives \begin{align*} &\frac{(x+(m+1)A) + (y+nB)}{2} \ge \sqrt{(x+mA) \cdot (y+(n+1)B)} \\ \iff& \frac{(x+mA) + (y+(n+1)B)}{2} \ge \sqrt{(x+mA) \cdot (y+(n+1)B)} + \frac{B-A}{2} \\ \iff& (x+mA) + (y+(n+1)B) - 2\sqrt{(x+mA) \cdot (y+(n+1)B)} \ge B-A \\ \iff& \left(\sqrt{x+mA} - \sqrt{y+(n+1)B}\right)^2 \ge B-A. \end{align*}Now assume for contradiction $0 < A < B$. Then any two terms of the sequences $(\sqrt{x+kA})_{k\ge 1}$ and $(\sqrt{y+kB})_{k \ge 2}$ need to differ by at least $\varepsilon = \sqrt{B-A} > 0$. However, these sequences are unbounded (as $A,B > 0$) and grow slower than linear (say, eventually consecutive terms differ by less than $\varepsilon$). This produces the required contradiction. $\blacksquare$
Assume now that $f$ is not the identity function; then by this point, we know there exists a single constant $D > 0$ such that \[ f(x) \in \left\{ x, x+D \right\} \]for every $x > 0$. We need to show in fact $f(x) = x+D$ for all $x$.
So suppose $f(x) = x+D$ and $f(y) = y$; let's see what we can deduce. In the case $x > y$, we use the estimate \[ P(x,y) \implies \frac{x+y+D}{2} \le \sqrt{\frac{x^2+y^2}{2}} < x \implies x > y+D. \]Meanwhile, in the case $x < y$ we instead use the estimate \[ Q(y,x) \implies \sqrt{y(x+D)} \le \frac{x+y}{2} < y \implies x < y-D. \]So combining $P(x,y)$ and $Q(y,x)$ is enough to prove the following statement: \[ f(x)=x+D \text{ and } f(y)=y \implies |x-y| > D. \qquad(\heartsuit) \]From $(\heartsuit)$ the problem is clearly solved, as its contrapositive says \[ f(x)=x+D \implies f(t) = t+D \; \forall t \in [x-D, x+D] \] Calculus solution. We briefly outline a calculus-based approach that does not use iterates. From the first solution, we have $f(f(x)) = 2f(x)-x$. Then $P(f(x), y)$ and $Q(f(x), y)$ together read \[ \sqrt{\frac{f(x)^2+f(y)^2}{2}} \ge f(x) + \frac{y-x}{2} \ge \sqrt{f(x)f(y)}. \]If we isolate $f(y)$ by direct brute-force calculation, the equation above becomes \[ \sqrt{f(x)^2 + 2(y-x)f(x) + \frac{(y-x)^2}{2}} \le f(y) \le f(x) + (y-x) + \frac{(y-x)^2}{4f(x)}. \]Treating $x$ and hence $f(x)$ as a fixed constant, the above inequality is good enough to imply that:
Claim: For each $x > 0$, we have \[ f(y) = f(x) + (y-x) + O(|y-x|^2) \qquad\text{as}\quad y \to x. \]Proof. [Sketch of proof] The upper bound is immediate. The lower bound is slightly more annoying due to the square root; but one may apply Taylor's theorem on the function \[ \sqrt{1+2t+t^2/2} = 1 + t - \frac{t^2}{4} + \frac{t^3}{4} - \frac{9t^4}{32} + \dots \]near $t = 0$; one then selects $t = (y-x) / f(x)$. $\blacksquare$
In particular, $f$ is differentiable and its derivative is $1$ everywhere; hence $f(x) \equiv x + c$ (and we need $c > 0$).

Solution 4 Solution4

Solution 5 Solution5