Let x and y be real numbers such that $6x^2 + 2xy + 6y^2 = 9 $.Find the maximum value of $x^2+y^2$ I tried to re-arrange the terms
$6x^2 + 2xy + 6y^2 = 9 $
$6x^2 + 6y^2 = 9 - 2xy $
$6 (x^2 + y^2) = 9 - 2xy $
$x^2 + y^2 = \frac{9 - 2xy}{6} $
Using A.M $\geq$ G.M
$\frac{x^2 + y^2}{2} \geq xy $
Can ayone help me from here? Am I going correct?

For AM-GM, $-2xy \le 2|xy| \le x^2+y^2$ so $2xy\ge -x^2-y^2$.
$$5(x^2+y^2)\le 6x^2+2xy+6y^2 =9.$$
Then $x^2+y^2\le 9/5$.  The maximum is reached when $x=-y=\pm\sqrt{9/10}$.
You can also get the min value of $x^2+y^2$.  Because $2xy\le 2|xy|\le x^2+y^2$
$$7(x^2+y^2)\ge 6x^2+2xy+6y^2=9.$$
Then $x^2+y^2\ge 9/7$.  The minimum is reached when $x=y=\pm\sqrt{9/14}$.
I think the shape $6x^2+2xy+6y^2=9$ is an ellipse with major/minor axes given by the lines $x=\pm y$. The axes have length $2\sqrt{9/5}$ and $2\sqrt{9/7}$.
