Bài 3: Thay $xy+xz+yz=1$ vào ta có:
VT = $\frac{x}{y\left ( x+y \right )\left ( x+z \right )}+\frac{y}{z\left ( y+x \right )\left ( y+z \right )}+\frac{z}{x\left ( z+x \right )\left ( z+y \right )}$
$= \frac{xyz\left ( x+y+z \right )+\left ( x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2} \right )}{xyz\left ( x+y \right )\left ( x+z \right )\left ( y+z \right )}$
$= \frac{1-xyz\left ( x+y+z \right )}{xyz\left ( x+y+z-xyz \right )}$
Ta cần CM:
$\frac{1-xyz\left ( x+y+z \right )}{xyz\left ( x+y+z-xyz \right )}\geq \frac{9}{4}\Leftrightarrow 9x^{2}y^{2}z^{2}+4\geq 13xyz\left ( x+y+z \right )$ (*)
Đặt $a= yz;b= xz;c=xy$ thì (*) $\Leftrightarrow 9abc+4\geq13\left ( ab+bc+ca \right )$
mà $a+b+c=1$, đến đây dồn biến ta có đpcm.