Question

Tìm GTNN của hàm số f(x)= 2x + \(\dfrac{8}{x^2}\) với  x \(\ge\) 4

Answer 1

\(\Rightarrow f\left(x\right)=\dfrac{7}{4}x+\dfrac{1}{8}x+\dfrac{1}{8}x+\dfrac{8}{x^2}\)

Áp dụng bđt Cô-si :

\(\dfrac{1}{8}x+\dfrac{1}{8}x+\dfrac{8}{x^2}\ge3\sqrt[3]{\dfrac{1}{8}x\cdot\dfrac{1}{8}x\cdot\dfrac{8}{x^2}}=\dfrac{3}{2}\)

\(\Rightarrow f\left(x\right)=\dfrac{7}{4}x+\dfrac{1}{8}x+\dfrac{1}{8}x+\dfrac{8}{x^2}\ge7+\dfrac{3}{2}=\dfrac{17}{2}\)

Dấu bằng xảy ra \(\Leftrightarrow x=4\)

Answer 2

\(f\left(x\right)=\dfrac{x}{8}+\dfrac{x}{8}+\dfrac{8}{x^2}+\dfrac{7}{4}x\ge3\sqrt[3]{\dfrac{8x^2}{64x^2}}+\dfrac{7}{4}.4=\dfrac{17}{2}\)

Dấu "=" xảy ra khi \(x=4\)