Track / Stage 5 / 365 of 400 #965 of 1964

Problem 965

AIME late

Algebra Difficulty 5.9 Prove it

Find, with proof, the largest possible value of

$\frac{x_{1}^{2}+\cdots+x_{n}^{2}}{n}$

where real numbers $x_{1}, \ldots, x_{n} \geq-1$ are satisfying $x_{1}^{3}+\cdots+x_{n}^{3}=0$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

For any $i$ , we have $0 \leq\left(x_{i}+1\right)\left(x_{i}-2\right)^{2}=x_{i}^{3}-3 x_{i}^{2}+4$ . Adding all of these we deduce that $\sum_{i=1}^{n} x_{i}^{2} \leq \frac{1}{3} \sum_{i=1}^{n}\left(x_{i}^{3}+4\right)=\frac{4}{3} n$ . Equality occurs, for example, when $n=9, x_{1}=\cdots=x_{8}=-1$ and $x_{9}=2$ . Therefore, the answer is $\frac{4}{3}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.