`text{Ta có:A=}` `x+1/x`
`=1/x+x/9+(8x)/9`
`text{Áp dụng bất đẳng thức Cô-si}`⇒$\frac{1}{x}$ +$\frac{x}{9}$ ≥2$\sqrt[]{\frac{1}{x}.\frac{x}{9}}$
`⇔1/x+x/9≥2√1/9`
`⇔1/x+x/9≥2/3`
$mà$ $\frac{8x}{9}$ ≥$\frac{8.3}{9}$ `text{(do x≥3)}`
`⇒`$\frac{8x}{9}$ ≥$\frac{8}{3}$
`⇔`$\left \{ {{\frac{1}{x}+\frac{x}{9}≥\frac{2}{3}} \atop {\frac{8x}{9}≥\frac{8}{9}}} \right.$
`⇔`$\frac{1}{x}$ +$\frac{x}{9}$ +$\frac{8x}{9}$ ≥$\frac{2}{3}$ +$\frac{8}{3}$
`⇔`$\frac{1}{x}$ +$\frac{x}{9}$ +$\frac{8x}{9}$ ≥$\frac{10}{3}$
`⇔A≥`$\frac{10}{3}$
`text{Dấu = xảy ra}`⇔$\left \{ {{\frac{1}{x}=\frac{x}{9}} \atop {x=3}} \right.$
`⇔x=3.`
`text{Vậy x=3 thì Min A=}` `10/3.`
