Cho $x,y,z>0$ . Chứng minh rằng:
$\frac{1}{x(x+y)}+\frac{1}{y(y+z)}+\frac{1}{z(z+x)}\geq \frac{27}{2(x+y+z)^{2}}$
Ta có : $(x+y)(y+z)(z+x)\leq \frac{(x+y+y+z+z+x)^3}{27}= \frac{8(x+y+z)^3}{27}$
$xyz\leq \frac{(x+y+z)^3}{27}$
$VT\geq \frac{3}{\sqrt[3]{xyz(x+y)(y+z)(z+x)}}\geq VP$
( Áp dụng 2 bđt ở trên )