1074

实数 ${\displaystyle x,y,z}$ 满足 ${\displaystyle xy+yz+zx=-1}$, 求 ${\displaystyle x^2+5y^2+8z^2}$ 的最小值.

注意到$$\begin{array}{rl}{x}^2+5y^2+8z^2-4=& x^2+5y^2+8z^2+4(xy+yz+zx{)}^2\\ =& (x+2y+2z{)}^2+(y-2z{)}^2\ge 0,\end{array}$$
因此 ${\displaystyle x^2+5y^2+8z^2\ge 4}$, 取等条件为 ${\displaystyle x=-6z,y=2z,z=\pm {\displaystyle \frac{1}{4}}}$.