Find the maximum and minimum of $x^2+2y^2$ if $x^2-xy+2y^2=1$. 
Find the maximum and minimum of $x^2+2y^2$ if $x,y\in\mathbb R$ and $$x^2-xy+2y^2=1$$

My attempt:
Clearly, since $x^2-x(y)+(2y^2-1)=0$ and $2y^2-y(x)+(x^2-1)=0$, we have that 
$$\Delta_1=y^2-8y^2+4=4-7y^2\ge 0$$ 
and
$$\Delta_2=x^2-8x^2+8=8-7x^2\ge 0$$
so that $y^2\le \frac{4}{7}$ and $x^2\le \frac{8}{7}$.
Thus $x^2+2y^2\le \frac{8}{7}+\frac{8}{7}=\frac{16}{7}$.
But clearly $x^2+2y^2\neq \frac{16}{7}$, since if $\begin{cases}x^2=\frac{8}{7}\\y^2=\frac{4}{7}\end{cases}$, then $x^2-xy+2y^2\neq 1$  
so that equality can't be achieved and we only have that $x^2+2y^2<\frac{16}{7}$.

An inequality approach: using $a^2+b^2\geq 2ab$ inequality below, we have
$$
x^2-xy+2y^2=1\implies 1+xy=x^2+2y^2\geq2\sqrt{2}xy\implies xy\leq\frac{1}{2\sqrt{2}-1}\cdot
$$
It follows that
$$
x^2+2y^2=1+xy\leq 1+\frac{1}{2\sqrt{2}-1}=
\boxed{\frac{2\sqrt{2}}{2\sqrt{2}-1}}\cdot
$$
Similarly, using $a^2+b^2\geq -2ab$, we have
$$
1+xy=x^2+2y^2\geq-2\sqrt{2}xy\implies xy\geq-\frac{1}{2\sqrt{2}+1}\cdot
$$
which implies
$$
x^2+2y^2=1+xy\geq 1-\frac{1}{2\sqrt{2}+1}=\boxed{\frac{2\sqrt{2}}{2\sqrt{2}+1}}\cdot
$$
Equality is realised for the max when
$$
y=\frac{1}{\sqrt{-\sqrt{2}+4}},\quad x=\sqrt{2}y;
$$
and for min
$$
y=\frac{1}{\sqrt{\sqrt{2}+4}},\quad x=-\sqrt{2}y\cdot
$$
