Track / Stage 5 / 105 of 400 #705 of 1964

Problem 705

AIME late

Algebra Difficulty 5.2 Prove it

Three. (20 points) Let $x, y, z \geqslant 0, x+y+z=3$ .
Prove: $\sqrt{x}+\sqrt{y}+\sqrt{z} \geqslant x y+y z+z x$ .

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

Three, because $2+(\sqrt{x})^{3}=1+1+(\sqrt{x})^{3} \geqslant 3 \sqrt{x}$ , so, $2 \sqrt{x}+x^{2} \geqslant 3 x$ .
Similarly, $2 \sqrt{y}+y^{2} \geqslant 3 y, 2 \sqrt{z}+z^{2} \geqslant 3 z$ .
Therefore, $2(\sqrt{x}+\sqrt{y}+\sqrt{z})+x^{2}+y^{2}+z^{2}$
$\geqslant 3(x+y+z)=(x+y+z)^{2}$ (since $x+y+z=3$ ).
Expanding yields $\sqrt{x}+\sqrt{y}+\sqrt{z} \geqslant x y+y z+z x$ .

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