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?
 A: Let $x= r\cos\theta , y= r\sin\theta$ , put it back into the given relation and get a function of $r^2 $in terms of $\theta$ we need to maximize this $r^ 2 $wrt to theta  ,use calculus or just bounds of trigo function
A: 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 $2\sqrt{9/7}$.
A: Without AM-GM inequality:
Let, $x=a+b$ and $y=a-b$, then you get
$$6(a+b)^2+2(a+b)(a-b)+6(a-b)^2=9$$
$$14a^2+10b^2=9$$
$$x^2+y^2=(a+b)^2+(a-b)^2=2a^2+2b^2$$
$$x^2+y^2=2a^2+2 \times \frac{9-14a^2}{10}$$
$$x^2+y^2=\frac{9-4a^2}{5}≤\frac95$$
$$\color {gold}{\boxed {\color{black} {\text{max}[x^2+y^2]=\frac 95.}}}$$

If you want to find the minimum value of the expression, you can write in the same way:
$$x^2+y^2=2a^2+2b^2$$
$$x^2+y^2=2b^2+2 \times \frac{9-10b^2}{14}$$
$$x^2+y^2=\frac{4b^2+9}{7}≥\frac 97$$
Then, you get
$$\color {gold}{\boxed {\color{black} {\text{min}[x^2+y^2]=\frac 97.}}}$$
A: 
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.
