Track / Stage 7 / 80 of 300 #1480 of 1964

Problem 1480

National olympiad second round; IMO P1/P4

Algebra Difficulty 7.2 Prove it

Note: This problem is from the 2007 Balkan Mathematical Olympiad, and it is almost identical to the 20th inequality in reference [3]: For positive real numbers $x, y, z$ satisfying the relation $x+y+z \geqslant x y + x z + y z$ , prove that $\frac{1}{1+x+y}+\frac{1}{1+y+z}+\frac{1}{1+z+x} \leqslant 1$ .

For the second problem: Let $x, y, z$ be non-negative numbers, and $x^{2}+y^{2}+z^{2}=3$ , prove that $\frac{x}{\sqrt{x^{2}+y+z}}+\frac{y}{\sqrt{y^{2}+z+x}}+\frac{z}{\sqrt{z^{2}+x+y}} \leqslant \sqrt{3}$ .

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

Proof: By the local Cauchy inequality, we have: $\left(x^{2}+y+z\right)(1+y+z) \geqslant(x+y+z)^{2}$ . Therefore, $\frac{1}{\sqrt{x^{2}+y+z}} \leqslant \frac{\sqrt{1+y+z}}{x+y+z}$ , and similarly, we can obtain the other two inequalities. Also, by the Cauchy inequality, $3\left(x^{2}+y^{2}+z^{2}\right) \geqslant(x+y+z)^{2}$ and $x^{2}+y^{2}+z^{2}=3$ imply: $x^{2}+y^{2}+z^{2} \geqslant x+y+z$ . Therefore, we need to prove $\frac{x \sqrt{1+y+z}+y \sqrt{1+z+x}+z \sqrt{1+x+y}}{x+y+z} \leqslant \sqrt{3}$ .

By the Cauchy inequality, we get $x \sqrt{1+y+z} + y \sqrt{1+z+x} + z \sqrt{1+x+y} \leqslant \sqrt{x+y+z} \sqrt{x+y+z+2xy+2xz+2yz} \leqslant \sqrt{x+y+z} \sqrt{x^{2}+y^{2}+z^{2}+2xy+2xz+2yz} = \sqrt{(x+y+z)^{3}}$ , i.e.,
$\begin{array}{l} \frac{x \sqrt{1+y+z}+y \sqrt{1+z+x}+z \sqrt{1+x+y}}{x+y+z} \leqslant \\ \sqrt{x+y+z} \leqslant \sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{3} \end{array}$

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