873

已知 $x^{2} - x y - 6 y^{2} = 1$ , 求 $x^{2} + y^{2} + x y$ 的最小值.

解钱莘芪

由条件 $\begin{array}{r} 1 = x^{2} - x y - 6 y^{2} = (x - 3 y) (x + 2 y), \end{array}$
记 $x - 3 y = t, x + 2 y = \frac{1}{t}$ ，解得 $x = \frac{1}{5} (2 t + \frac{3}{t}), y = \frac{1}{5} (\frac{1}{t} - t)$ .这样 $\begin{array}{r} X^{2} + y^{2} + x y = \frac{1}{25} (3 t^{2} + \frac{13}{t^{2}} + 9) \geq \frac{2 \sqrt{39} + 9}{25} . \end{array}$