ＭＴ【106】待定系数

解答：$\because(\frac{\sqrt{5}}{2}x^2+\frac{1}{2\sqrt{5}}y^2)+(\frac{2}{\sqrt{5}}y^2+\frac{\sqrt{5}}{2}z^2)\ge xy+2yz\therefore$最大值为$\frac{\sqrt{5}}{2}$