Giải thích các bước giải:
Ta có :
$\sqrt{\dfrac{xy}{xy+z}}$
$=\sqrt{\dfrac{xy}{xy+z(x+y+z)}}$
$=\sqrt{\dfrac{xy}{(z+x)(z+y)}}$
$=\sqrt{\dfrac{x}{x+z}}.\sqrt{\dfrac{y}{y+z}}\le\dfrac12(\dfrac{x}{x+z}+\dfrac{y}{y+z})$
Tương tự
$\sqrt{\dfrac{yz}{yz+x}}\le\dfrac12(\dfrac{y}{x+y}+\dfrac{z}{x+z})$
$\sqrt{\dfrac{zx}{zx+y}}\le\dfrac12(\dfrac{z}{z+y}+\dfrac{x}{y+x})$
Cộng vế với vế
$\to \sqrt{\dfrac{xy}{xy+z}}+\sqrt{\dfrac{yz}{yz+x}}+\sqrt{\dfrac{zx}{zx+y}}\le\dfrac12(\dfrac{x}{x+z}+\dfrac{y}{y+z})+\dfrac12(\dfrac{y}{x+y}+\dfrac{z}{x+z})+\dfrac12(\dfrac{z}{z+y}+\dfrac{x}{y+x})$
$\to \sqrt{\dfrac{xy}{xy+z}}+\sqrt{\dfrac{yz}{yz+x}}+\sqrt{\dfrac{zx}{zx+y}}\le\dfrac32$
Dấu = xảy ra khi $x=y=z=\dfrac13$
