, 2014

Solution:
Answer: $\frac{72+32 \sqrt{2}}{7}$

Let $u = x^{2} + 2 y^{2}$. By AM-GM, $u \geq \sqrt{8} x y$, so $x y \leq \frac{u}{\sqrt{8}}$. If we let $x y = k u$ where $k \leq \frac{1}{\sqrt{8}}$, then we have

$$\begin{gathered} u(1-k)=8 \\ u(1+k)=x^{2}+x y+2 y^{2} \end{gathered}$$

that is, $u(1+k)=8 \cdot \frac{1+k}{1-k}$. It is not hard to see that the maximum value of this expression occurs at $k=\frac{1}{\sqrt{8}}$, so the maximum value is $8 \cdot \frac{1+\frac{1}{\sqrt{8}}}{1-\frac{1}{\sqrt{8}}}=\frac{72+32 \sqrt{2}}{7}$.