\(已知：x,y \geq 0,且x+3y=x^3y^2,\)
\(求：\frac{3}{x}+\frac{2}{y}的最小值。\)
方法一，老师教的：
\[
\begin{align*}
(\frac{3}{x}+\frac{2}{y})^2 \\
& = \frac{9}{x^2}+\frac{12}{xy} +\frac{4}{y^2} \\
& = \frac{9}{x^2} + \frac{12y+4x}{xy^2} \\
& = \frac{9}{x^2} + \frac{4(x+3y)}{xy^2} \\
& = \frac{9}{x^2} + \frac{4x^3y^2}{xy^2} \\
& = \frac{9}{x^2} + 4x^2 \\
由公式：a^2+b^2 \geq 2ab:
& \geq 2 \cdot (\frac{3}{x} \cdot 2x) \\
& \geq 2 \cdot (6) \\
& \geq 12 \\
即解得 \frac{3}{x} + \frac{2}{y} 最小值为:\sqrt{12} = 2\sqrt{3}.
\end{align*}
\]
方法二，学生做的：
\[
\begin{align*}
& x+3y = x^3y^2 \\
& x(\frac{1}{y})^2 + 3(\frac{1}{y}) -x^3 = 0 \\
& (\frac{1}{y}) =\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\
& = \frac{-3 \pm \sqrt{3^2-4x \cdot (-x^3)}}{2x} \\
& = \frac{-3 \pm \sqrt{9+4x^4}}{2x} \\
\frac{3}{x}+\frac{2}{y} \\
& = \frac{3}{x} + 2(\frac{-3 \pm \sqrt{9+4x^4}}{2x}) \\
& = \frac{\pm \sqrt{9+4x^4}}{x} \\
已知x,y \geq 0 , 负数舍去。 \\
& = \frac{\sqrt{9+4x^4}}{x} \\
& = \sqrt{\frac{9}{x^2}+\frac{4x^4}{x^2}} \\
& = \sqrt{\frac{9}{x^2}+\frac{4x^2}{1}} \\
由公式：a^2+b^2 \geq 2ab : \\
& \geq \sqrt{2(\frac{3}{x} \cdot \frac{2x}{1})} \\
& \geq 2\sqrt{3}
即原式\frac{3}{x}+\frac{2}{y}的最小值为：2\sqrt{3}.
\end{align*}
\]