已知 \(x,y,z\in \mathbb{R}^+\)，且\(x+4y+3z=6\)，求\(x^2yz^3\)的最大值。
解﹕\(\because x+4y+3z\)
\(=\frac{x}{2}+\frac{x}{2}+4y+z+z+z\)
\(\geq 6\sqrt[6]{\frac{x}{2}\cdot\frac{x}{2}\cdot 4y\cdot z\cdot z\cdot z}=6\sqrt[6]{x^2 y z^3}\)
\(\therefore x+4y+3z\geq 6\sqrt[6]{x^2 y z^3}\)
\(\therefore 6\geq 6\sqrt[6]{x^2 y z^3}\)
\(\therefore x^2 y z^3\leq 1\)
當且僅當 \(\frac{x}{2}=4y=z=1\)，即\(x=2,y=\frac{1}{4},z=1\)時，\(x^2 y z^3\) 的最大值是 1