840

若 $x, y, z > 0$ 且 $2 x + 2 y + z = 1$ , 求 $3 x y + y z + z x$ 的最大值.

解钱莘芪

直接消去 $z$ 得 $\begin{aligned} 3 x y + z (x + y) & = 3 x y - 2 (x + y)^{2} + (x + y) \\ \leq - \frac{5}{4} (x + y)^{2} + (x + y) = \frac{1}{5} - {(\frac{\sqrt{5}}{2} (x + y) - \frac{1}{\sqrt{5}})}^{2} \leq \frac{1}{5} . \end{aligned}$
取等时 $(x, y, z) = (\frac{1}{5}, \frac{1}{5}, \frac{1}{5})$ .