INAMO 2024/7 wrote: Suppose $P(x) = x^n + a_{n-1} x^{n-1} + \cdots + a_1x + a_0$ where $a_0, a_1, \ldots, a_{n-1}$ are reals for $n\geq 1$ (monic $n$th-degree polynomial with real coefficients). If the inequality \[ 3(P(x)+P(y)) \geq P(x+y) \]holds for all reals $x,y$, determine the minimum possible value of $P(2024)$. This dies to quick "analysis" tricks. There might be some miscalculation but I think the general idea should work fine: Claim 01. $\text{deg}(P) = 2$, and thus we can write $P$ as $P(x) = (x + \alpha)^2 + \beta$ for some $\alpha, \beta \in \mathbb{R}$. Proof. Note that we have $P(2x) \le 6P(x)$ for all $x \in \mathbb{R}$, i.e. $\frac{P(2x)}{P(x)} \le 6$ for all $x$ such that $P(x) \not= 0$. As $\text{deg}(P) \ge 1$, we have $2^{\text{deg}(P)} = \lim_{x \to \infty} \frac{P(2x)}{P(x)} \le 6$, and thus $\text{deg}(P) \le 2$. As $P(x) \ge -\frac{3}{2} P(0)$ for all $x \in\mathbb{R}$, i.e. $P$ is bounded from below, then $P$ can't be linear and thus $P(x) = (x + \alpha)^2 + \beta$ for some $\alpha, \beta \in \mathbb{R}$. The problem condition thus translates to $3((x + \alpha)^2 + (y + \alpha)^2) + 5 \beta \ge (x + y + \alpha)^2$ holds for all $x,y \in \mathbb{R}$. Fix $\alpha \in \mathbb{R}$. Claim 02. $\beta \ge \frac{3}{5} \alpha^2$ and this is tight. Proof. To see this, just note that when $x = y = -2 \alpha$, we have obtain the desired lower bound $\beta \ge \frac{3}{5} \alpha^2$. To see that $\beta = \frac{3}{5}\alpha^2$ works, we see that for any $x,y \in \mathbb{R}$, \begin{align*} 3((x + \alpha)^2 + (y + \alpha)^2 + \alpha^2) - (x + y + \alpha)^2 &= \frac{1}{2} (2 \alpha + 2x - y)^2 + \frac{3}{2} (2\alpha + y)^2 \ge 0 \end{align*} To finish this, we are asked to minimize $P(2024) = (2024 + \alpha)^2 + \beta \ge (2024 + \alpha)^2 + \frac{3}{5} \alpha^2 \ge 1536216$ where equality holds when $\alpha = -1265$ and $\beta = 960135$.

I got $P(x) \ge -3P(0)/2$ which implies $P(x)$ bounded below, happen only when $\deg P$ even. For $\deg P \ge 4$ and look at the leading coefficient $3P(x) \ge P(2x)$, take $x \to \infty$ lead contradiction Next argument similar with above.