HMMT-Feb 2014 guts, Problem 16

Suppose that $x$ and $y$ are positive real numbers such that $x^{2}-x y+2 y^{2}=8$. Find the maximum possible value of $x^{2}+x y+2 y^{2}$.

Solution: 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 $8 \cdot \frac{1+\frac{1}{\sqrt{8}}}{1-\frac{1}{\sqrt{8}}}=\frac{72+32 \sqrt{2}}{7}$. $\fbox{\frac{72+32 \sqrt{2}}{7}}$.

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