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?


Let set: $\begin{cases}f(x,y)=x^2+y^2\\g(x,y)=6x^2+2xy+6y^2\end{cases}$
The curve $(\mathcal E): g(x,y)=9$ is an ellipse centered at the origin with major axis supported by $y=-x$ and minor axis supported by $y=x$.
Indeed if you calculate its equation in the $45^\circ$ rotated basis then you get it in its reduced form:
$$g(\tfrac{X+Y}{\sqrt{2}},\tfrac{X-Y}{\sqrt{2}})=7X^2+5Y^2=9$$
On the other hand, the curve $(\mathcal C): f(x,y)=k$ is just a circle centered at origin.
The linked extrema occurs when this circle is tangent to the ellipse, therefore the minimum occurs on the minor axis (i.e. when $x=y$) and the maximum occurs on the major axis (i.e. when $x=-y$).
$\begin{cases}g(x,x)=14x^2=9\\g(x,-x)=10x^2=9\end{cases}\implies \begin{cases}f(x,x)=2x^2=\frac 97&\text{min}\\f(x,-x)=2x^2=\frac 95&\text{max}\end{cases}$
Note: you can also extrapolate these values directly from the reduced equation of the ellipse.
