Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$
(x-y)(f(x)+f(y)) \leqslant f\left(x^{2}-y^{2}\right) \quad \text { for all } x, y \in \mathbb{R}
$$

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Clearly, $f(x)=c x$ is a solution for each $c \in \mathbb{R}$ since $(x-y)(c x+c y)=c\left(x^{2}-y^{2}\right)$. To show that there are no other solutions, we observe that

(1) $x=$

$$
x=1, y=0: \quad f(0)+f(1) \leqslant f(1) \Rightarrow f(0) \leqslant 0 \text {, whence } f(0)=0 \text {; }
$$

(2) $y=-x: \quad 2 x(f(x)+f(-x)) \leqslant f(0)=0$;

$$
x \rightarrow-x: \quad-2 x(f(-x)+f(x)) \leqslant 0 \Rightarrow 2 x(f(x)+f(-x)) \geqslant 0 ;
$$

thus $2 x(f(x)+f(-x))=0$ for all $x$, so $f(-x)=-f(x)$ for all $x \neq 0$, and hence for all $x$, since $f(0)=0$;

(3) $x \leftrightarrow y: \quad(y-x)(f(y)+f(x)) \leqslant f\left(y^{2}-x^{2}\right)=-f\left(x^{2}-y^{2}\right) \Rightarrow(x-y)(f(x)+f(y)) \geqslant f\left(x^{2}-y^{2}\right)$; which is the given inequality with the inequality sign reversed, so $(x-y)(f(x)+f(y))=f\left(x^{2}-y^{2}\right)$ must hold for all $x, y \in \mathbb{R}$;

(4) $y \leftrightarrow-y: \quad(x-y)(f(x)+f(y))=f\left(x^{2}-y^{2}\right)=f\left(x^{2}-(-y)^{2}\right)=(x+y)(f(x)+f(-y))=(x+y)(f(x)-f(y))$; expanding yields $f(x) y=f(y) x$ for all $x, y \in \mathbb{R}$.

Taking $y=1$ in the last result, $f(x)=f(1) x$, i.e. $f(x)=c x$, where $c=f(1)$, for all $x \in \mathbb{R}$. Since we have shown above that, conversely, all such functions are solutions, this completes the proof.

Alternative solution. A slight variation of this argument proves that $(x-y)(f(x)+f(y))=f\left(x^{2}-y^{2}\right)$ must hold for all $x, y \in \mathbb{R}$ as above, and then reaches $f(x)=c x$ as follows:

(4) $y= \pm 1: \quad(x \mp 1)(f(x) \pm f(1))=f\left(x^{2}-1\right)$ using $f(-1)=-f(1)$ from (2);

hence $(x-1)(f(x)+f(1))=(x+1)(f(x)-f(1)) \Rightarrow f(x)=f(1) x=c x$, where $c=f(1)$, on expanding.

## BxMO 2023: Problems and Solutions

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