\(A=\dfrac{x}{9}+\dfrac{1}{x}+\dfrac{8x}{9}\)
Áp dụng BĐT Cô-si: \(\dfrac{x}{9}+\dfrac{1}{x}\ge2\sqrt{\dfrac{x}{9x}}=\dfrac{2}{3}\)
Đồng thời do \(x\ge3\Rightarrow\dfrac{8x}{9}\ge\dfrac{8.3}{9}=\dfrac{8}{3}\)
\(\Rightarrow A\ge\dfrac{2}{3}+\dfrac{8}{3}=\dfrac{10}{3}\)
\(A_{min}=\dfrac{10}{3}\) khi \(x=3\)