H
hien_vuthithanh


cho x,y,z là 3 số dương thoả mãn x+y+z=3.CM
$\dfrac{x}{x+\sqrt{3x+yz}}+\dfrac{y}{y+\sqrt{3y+zx}}+\dfrac{z}{z+\sqrt{3z+xy}}$ \leq 1
$\dfrac{x}{x+\sqrt{3x+yz}}+\dfrac{y}{y+\sqrt{3y+zx}}+\dfrac{z}{z+\sqrt{3z+xy}}$ \leq 1