Question

Cho x,y,z>0 x+y+z=1. CMR \(x^2+y^2+z^2+2\sqrt{3xyz}\le1\)

Answer 1

Lời giải:
BĐT $\Leftrightarrow (x+y+z)^2-2(xy+yz+xz)+2\sqrt{3xyz}\leq 1$

$\Leftrightarrow 1-2(xy+yz+xz)+2\sqrt{3xyz}\leq 1$

$\Leftrightarrow xy+yz+xz\geq \sqrt{3xyz}$

$\Leftrightarrow (xy+yz+xz)^2\geq 3xyz=3xyz(x+y+z)$

$\Leftrightarrow (xy+yz+xz)^2-3xyz(x+y+z)\geq 0$

$\Leftrightarrow (xy-yz)^2+(yz-xz)^2+(xz-xy)^2\geq 0$ (luôn đúng với mọi $x,y,z>0$)

Ta có đpcm

Dấu "=" xảy ra khi $x=y=z=\frac{1}{3}$