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How to aproach the following problem:

Find smallest and biggest values of the expression: \begin{align*} x^2+2y^2 \end{align*} When \begin{align*} x^2-xy+2y^2 &= 1 \end{align*}

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    lagrange optimization? – 2017-01-28
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    This problem was originally posted at math competition for 11 graders. I think that the solution of this problem does not require any derivatives or other fancy stuff. – 2017-01-28
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    Just to note, if $|x| \leq 1$ and $|y| \leq 1$, then it is easy to see that the max and min values of $x^2 + 2y^2$ are $1$ and $2$ respectively. – 2017-01-28
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    Nilabro, $(x-1/2y)^2+7/4y^2=1 >> \abs (y) \le \sqrt{4/7}$ Same can be done with $x$ – 2017-01-28

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Let $x^2+2y^2=k$. Note that from our conditions $k-1=x^2+2y^2-1=xy$. Because of $\text{AM-GM}$ we have that $$k=x^2+2y^2 \ge 2 \sqrt{2}xy =2\sqrt{2}(k-1)$$ Which gives us the maximum with equality when $x=\sqrt{2}y$. Similarly, $$k=x^2+2y^2 \ge -2\sqrt{2}xy=-2\sqrt{2}(k-1)$$ Gives us the minimum with equality when $x=-\sqrt{2}y$. Thus we have $$\frac{8+2\sqrt{2}}{7} \ge x^2+2y^2 \ge \frac{8-2\sqrt{2}}{7}$$ We are done.