[tex]\frac{1}{x^{2}+x}+\frac{x}{2}+\frac{x+1}{4}\geq 3\sqrt[3]{\left ( \frac{1}{x^{2}+x}\cdot \frac{x}{2}\cdot \frac{x+1}{4} \right )}=3\sqrt[3]{\frac{1}{8}}=\frac{3}{2}[/tex]
Tương tự,
[tex]\frac{1}{y^{2}+y}+\frac{y}{2}+\frac{y+1}{4}\geq \frac{3}{2}[/tex]
[tex]\frac{1}{z^{2}+y}+\frac{z}{2}+\frac{z+1}{4}\geq \frac{3}{2}[/tex]
[tex]\Rightarrow \frac{1}{x^{2}+x}+\frac{1}{y^{2}+y}+\frac{1}{z^{2}+z}+\frac{x+y+z}{2}+\frac{3+x+y+z}{6}\geq \frac{9}{2}[/tex]
[tex]\Leftrightarrow \frac{1}{x^{2}+x}+\frac{1}{y^{2}+y}+\frac{1}{z^{2}+z}+\frac{3}{2}+\frac{3}{2} \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]