Có :
[tex]\frac{1}{x(x+1)}+\frac{x}{2}+\frac{x+1}{4}\geq 3 \sqrt[3]{\frac{1}{x(x+1)}.\frac{x}{2}.\frac{x+1}{4}}=\frac{3}{2}[/tex] (1)
[tex]\frac{1}{y(y+1)}+\frac{y}{2}+\frac{y+1}{4}\geq 3 \sqrt[3]{\frac{1}{y(y+1)}.\frac{y}{2}.\frac{y+1}{4}}=\frac{3}{2}[/tex] (2)
[tex]\frac{1}{z(z+1)}+\frac{z}{2}+\frac{z+1}{4}\geq 3 \sqrt[3]{\frac{1}{z(z+1)}.\frac{z}{2}.\frac{z+1}{4}}=\frac{3}{2}[/tex] (3)
Cộng (1), (2), (3) theo vế ta có :
[tex]\frac{1}{x(x+1)}+\frac{1}{y(y+1)}+\frac{1}{z(z+1)}+\frac{x+y+z}{2}+\frac{x+y+z+3}{4}\geq \frac{9}{2}[/tex] [tex]\Leftrightarrow \frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}\geq \frac{3}{2}[/tex]