<s>[INST] maximum and minimum value of $\frac{x^2+y^2}{x^2+xy+4y^2}$ 
If $x,y\in\mathbb{R}$ and $x^2+y^2>0.$ Then maximum and minimum value of $\displaystyle \frac{x^2+y^2}{x^2+xy+4y^2}$

Plan
Let $$K=\frac{x^2+y^2}{x^2+xy+4y^2}$$
$$Kx^2+Kxy+4Ky^2=x^2+y^2\Rightarrow (4K-1)y^2+Kxy+(K-1)x^2=0$$
put $y/x=t$ and equation is $(4K-1)t^2+Kt+(K-1)=0$
How do i solve it Help me plesse
 [/INST] Let $v = (x,y)$,
$$
A = \begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 4 \end{bmatrix}
$$
and define the Rayleigh cuocient.
$$
R_A(v) = \frac{v^T A v}{v^Tv}.
$$
Then, it is clear that
$$
\frac{x^2+y^2}{x^2+xy+4y^2} = \frac{1}{R_A(v)}.
$$
It is well-known (Linear Algebra result) that $\operatorname{min}_{v \neq 0} R_A(v) = \lambda_{\operatorname{min}}$ and $\operatorname{max}_{v \neq 0} R_A(v) = \lambda_{\operatorname{max}}$, where $\lambda_{\operatorname{min}}$ and $\lambda_{\operatorname{max}}$ are the smallest and largest eigenvalues of $A$, respectively.
In this case $\lambda_{\operatorname{min}} = \frac{1}{2} \left(5-\sqrt{10}\right)$ and $\lambda_{\operatorname{max}} = \frac{1}{2} \left(5+\sqrt{10}\right)$. Therefore,
$$
\min_{x,y\neq 0} \frac{x^2+y^2}{x^2+xy+4y^2} = \frac{1}{\max_{v\neq 0} R_A(v)} = \frac{1}{\frac{1}{2} \left(5+\sqrt{10}\right)} = \frac{2}{15}(5+\sqrt{10})
$$
and
$$
\max_{x,y\neq 0} \frac{x^2+y^2}{x^2+xy+4y^2} = \frac{1}{\min_{v\neq 0} R_A(v)} = \frac{1}{\frac{1}{2} \left(5-\sqrt{10}\right)} = \frac{2}{15}(5-\sqrt{10}).
$$
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