Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.9 AIME, harder Prove it

Example 9.7 (1991 Polish Mathematical Olympiad) $x^{2}+y^{2}+z^{2}=2$ , prove that
$x+y+z \leqslant x y z+2$

Solution

Assume $x \leqslant y \leqslant z$ , then $x y \leqslant 1, (x-1)(y-1) \geqslant 0$ .
If $z \geqslant 1$ , then
$x+y+z \leqslant x y z+2 \Leftrightarrow 2(x-1)(y-1)(z-1)+(x+y+z-2)^{2} \geqslant 0$

If $z<1$ , then
$x+y+z \leqslant x y z+2 \Leftrightarrow(x-1)(y-1)+(z-1)(x y-1) \geqslant 0$

Combining both cases, the inequality is proved!

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