\(x\in \mathbb{R}^+\),求下列各式的最小值.
(1) \(x^2+\frac{3}{x}\)
解﹕\(\because x>0\)
\(x^2+\frac{3}{x}=x^2+\frac{3}{2x}+\frac{3}{2x}\geq 3\sqrt[3]{x^2\cdot\frac{3}{2x}\cdot\frac{3}{2x}}=\frac{3}{2}\sqrt[3]{18}\)
當且僅當 \(x^2=\frac{3}{2x}\),即\(x=\frac{\sqrt[3]{12}}{2}\)時，\(x^2+\frac{3}{x}\)的最小值是 \(\frac{3}{2}\sqrt[3]{18}\)
(2) \(x+\frac{3}{x^2}\)
解﹕\(\because x>0\)
\(x+\frac{3}{x^2}=\frac{x}{2}+\frac{x}{2}+\frac{3}{x^2}\geq 3\sqrt[3]{\frac{x}{2}\cdot\frac{x}{2}\cdot\frac{3}{x^2}}=\frac{3}{2}\sqrt[3]{6}\)
當且僅當 \(\frac{x}{2}=\frac{3}{x^2}\),即\(x=\sqrt[3]{6}\)時，\(x+\frac{3}{x^2}\)的最小值是\(\frac{3}{2}\sqrt[3]{6}\)