Find the points on the curve $x^4+y^4+3xy=2$ closest and farthest to the origin I need to find the points on the curve $x^4+y^4+3xy=2$ that are closest and farthest to the origin. 
I believe this might be a Lagrange multiplier problem, but I am not sure. I was thinking that maybe minimizing/maximizing the function would be the way to go.

We'll prove that $\max(x^2+y^2)=4.$
Indeed, let $xy\geq0$.
Thus, by C-S $$2=x^4+y^4+3xy\geq x^4+y^4=\frac{(1+1)(x^4+y^4)}{2}\geq\frac{(x^2+y^2)^2}{2},$$
which gives $$x^2+y^2\leq2<4.$$
Now, let $xy\leq0.$
Thus, by C-S again and AM-GM we obtain:
$$2=x^4+y^4+3xy\geq\frac{(x^2+y^2)^2}{2}-\frac{3(x^2+y^2)}{2},$$
which gives $$(x^2+y^2)^2-3(x^2+y^2)-4\leq0$$ or
$$(x^2+y^2-4)(x^2+y^2+1)\leq0$$ or
$$x^2+y^2\leq4.$$
The equality occurs for $(x^2,y^2)||(1,1)$ and $x=-y,$ which with our condition gives:
$$(x,y)=(\sqrt2,-\sqrt2)$$ or
$$(x,y)=(-\sqrt2,\sqrt2).$$
By the similar way we can prove that $\min(x^2+y^2)=1,$ where the equality occurs for $$(x,y)=\left(-\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right)$$ or 
$$(x,y)=\left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\right).$$ 
