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Berkeley Math Circle: Monthly Contest 2

Algebra Difficulty 5.0 AIME Prove it United States

Problem:
Find, with proof, the largest possible value of

x12+⋯+xn2n \frac{x_{1}^{2}+\cdots+x_{n}^{2}}{n}

where real numbers x1,…,xn≥−1x_{1}, \ldots, x_{n} \geq -1 are satisfying x13+⋯+xn3=0x_{1}^{3}+\cdots+x_{n}^{3}=0.

Solution

Solution:
For any ii, we have 0≤(xi+1)(xi−2)2=xi3−3xi2+40 \leq (x_{i}+1)(x_{i}-2)^{2} = x_{i}^{3} - 3x_{i}^{2} + 4. Adding all of these we deduce that
∑i=1nxi2≤13∑i=1n(xi3+4)=43n. \sum_{i=1}^{n} x_{i}^{2} \leq \frac{1}{3} \sum_{i=1}^{n} (x_{i}^{3} + 4) = \frac{4}{3} n.
Equality occurs, for example, when n=9n=9, x1=⋯=x8=−1x_{1}=\cdots=x_{8}=-1 and x9=2x_{9}=2. Therefore, the answer is 43\frac{4}{3}.

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