đề 1
Ta có:
$2P=\frac{2\sqrt{yz}}{x+2\sqrt{yz}}+\frac{2\sqrt{zx}}{y+2\sqrt{zx}}+\frac{2\sqrt{xy}}{z+2\sqrt{xy}}$
$=1-\frac{x}{x+2\sqrt{yz}}+1-\frac{y}{y+2\sqrt{xz}}+1-\frac{z}{z+2\sqrt{xy}}$
$=3-(\frac{\sqrt{x}^2}{x+2\sqrt{yz}}+\frac{\sqrt{y}^2}{y+2\sqrt{xz}}+\frac{\sqrt{z}^2}{z+2\sqrt{xy}})$
$\leq 3-\frac{({\sqrt{x}+\sqrt{y}+\sqrt{z}})^2}{x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{zx}}$ (Cauchy-Schwarz)
$=3-\frac{({\sqrt{x}+\sqrt{y}+\sqrt{z}})^2}{({\sqrt{x}+\sqrt{y}+\sqrt{z}})^2}$
$=3-1=2$
Suy ra $P \leq 1$
Vậy, maxP=1
Dấu bằng xảy ra khi x=y=z