Exercise 12. Find all polynomials $\mathrm{P}$ with real coefficients such that for all real numbers $x, y$,

$$
x P(x) + y P(y) \geqslant 2 P(x y)
$$
Solution to Exercise 12 If $P(X) = cX$ with $c \geq 0$, then $xP(x) + yP(y) = c(x^2 + y^2) \geq 2cxy = 2P(xy)$ by the mean inequality. Polynomials of the form $cX$ with $c \geq 0$ are suitable; let's show that these are the only ones. Suppose $P$ is non-constant. By evaluating the inequality at $(x, 0)$, we get $xP(x) \geq 2P(0)$. By considering the limit as $x \to +\infty$, we see that the leading coefficient of $P$ must be strictly positive (since $P$ cannot tend to $-\infty$). By evaluating at $(x, x)$, we get $xP(x) \geq P(x^2)$. The polynomial $XP(X)$ has degree $\deg(P) + 1$, and $P(X^2)$ has degree $2\deg(P)$. If $2\deg(P) > \deg(P) + 1$, then $P(X^2) - XP(X)$ is a polynomial of degree $2\deg(P)$ with a strictly positive leading coefficient, so it is strictly positive for sufficiently large $x$, leading to a contradiction. Therefore, $2\deg(P) \leq \deg(P) + 1$, so $P$ is a polynomial of degree at most 1 with a strictly positive leading coefficient.

Now let $P(X) = aX + b$ with $a \geq 0$. By evaluating the inequality at $(x, x)$, we get $ax^2 + bx \geq ax^2 + b$, so $bx \geq b$. For $x = 2$, we get $b \geq 0$, and for $x = 0$, we get $b \leq 0$, so $b = 0$. Thus, $P(X) = aX$ with $a \geq 0$, which concludes the proof.

Comment from the graders: Students often assumed that the leading coefficient was positive, but it needs to be verified at the beginning. Similarly, the degree of the polynomial must be justified. Finally, it is important not to forget to verify that the polynomials found satisfy the statement of the problem.