Question

cho x>0 tìm giá trị nhỏ nhất \(x+\dfrac{32}{x^2}\)

Answer 1

\(x+\dfrac{32}{x^2}=\dfrac{x}{2}+\dfrac{x}{2}+\dfrac{32}{x^2}\ge3\sqrt[3]{\dfrac{x}{2}.\dfrac{x}{2}.\dfrac{32}{x^2}}=3\sqrt[3]{\dfrac{32}{4}}=6\)

\(Min=6\Leftrightarrow\dfrac{x}{2}=\dfrac{32}{x^2}\Leftrightarrow x^3=64\Leftrightarrow x=4\)

Answer 2

\(\Leftrightarrow x+\dfrac{\left(4\sqrt{2}\right)^2}{x^2}\Leftrightarrow x+\dfrac{4\sqrt{2}}{x}\)

ta có x>0

áp dụng BĐT Cô si ta có:

\(x+\dfrac{4\sqrt{2}}{x}\ge2\sqrt{x.\dfrac{4\sqrt{2}}{x}}\)

\(\Leftrightarrow x+\dfrac{4\sqrt{2}}{x}\ge2\sqrt{4\sqrt{2\simeq}4,75}\)

dấu = xảy ra khi x\(\simeq2,37\)