Chứng minh rằng nếu x, y, z là số dương thì $\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge 9$ .

Ta có: $\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{x}{x}+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{y}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+\frac{z}{z}$

$=\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+3=3+\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)$

Áp dụng bất đẳng thức Cô – si ta có

$\frac{x}{y}+\frac{y}{x}\ge 2\sqrt{\frac{x}{y}.\frac{y}{x}}=2$

$\frac{x}{z}+\frac{z}{x}\ge 2\sqrt{\frac{x}{z}.\frac{z}{x}}=2$

$\frac{y}{z}+\frac{z}{y}\ge 2\sqrt{\frac{y}{z}.\frac{z}{y}}=2$

Suy ra $\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)\ge 2+2+2=6$

Do đó $\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge 3+6=9$

Vậy $\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge 9$ .