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Problem 894

AIME late

Algebra Difficulty 5.7 Prove it

Question 7. Let $x, y, z$ be positive real numbers satisfying $x, y, z < 2$ and $x^{2} + y^{2} + z^{2} = 3$ . Prove:
$\frac{3}{2} < \frac{1 + y^{2}}{x + 2} + \frac{1 + z^{2}}{y + 2} + \frac{1 + x^{2}}{z + 2} < 3.$
(2008 Greek Mathematical Olympiad)
The left inequality can be strengthened to:
$\frac{1 + y^{2}}{x + 2} + \frac{1 + z^{2}}{y + 2} + \frac{1 + x^{2}}{z + 2} \geq 2.$

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Proof: Given that $x, y, z \leq \sqrt{3}$ , by the Cauchy-Schwarz inequality and the weighted power mean inequality, we have
$\begin{array}{c} \sum \frac{x^{2}+1}{z+2} \geq \sum \frac{x^{2}+1}{\frac{z^{2}+1}{2}+2}=2 \sum \frac{x^{2}+1}{z^{2}+5} \geq \frac{2\left(x^{2}+y^{2}+z^{2}+3\right)^{2}}{\sum\left(x^{2}+1\right)\left(z^{2}+5\right)} \\ =\frac{72}{\sum x^{2} y^{2}+33} \geq \frac{72}{\frac{1}{3}\left(x^{2}+y^{2}+z^{2}\right)^{2}+33}=\frac{72}{36}=2 . \end{array}$

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