Find a minimum of $x^2+y^2$ under the condition $x^3+3xy+y^3=1$ As in the title, I've tried to find a maximum and mininum of $x^2+y^2$ when $x^3+3xy+y^3=1$ holds. It is not too hard to show that $x^2+y^2$ has no maximum, but I can't find a minimum.
Lagrange multiplier gives a dirty calculation so I can't handle it. Is there any elegant way to find it? Thanks for any help.
p.s. Sorry. I make a typo in the $xy$-coefficient.

Noting that
$$x^3+y^3=(x+y)^3-3xy(x+y)$$
and
$$(x+y)^3-1=(x+y-1)\left[(x+y)^2+(x+y)+1\right]$$
gives
$$x^3+3xy+y^3=1 \\\\ \implies\ (x+y)^3-1-3xy(x+y)+3xy=0 \\\\ \implies\ (x+y-1)\left[(x+y)^2+(x+y)+1-3xy\right]=0 \\\\ \implies\ (x+y-1)(x^2+y^2-xy+x+y+1)=0$$
So either $x+y=1$ or $x^2+y^2-xy+x+y+1=0$.
Treating the latter as a quadratic equation in $x$ gives its discriminant as $(y-1)^2-4(y^2+y+1)=-3(y+1)^2$; so a real solution exists only when $y=-1$, at which value $x=-1$. This gives $x^2+y^2=(-1)^2+(-1)^2=2$.
As for $x+y=1$, we want the circle of $x^2+y^2=r$ $(r\ge0)$ to be tangent to that straight line. This occurs at $x=y=\dfrac12$, giving $r_{\min}=\left(\dfrac12\right)^2+\left(\dfrac12\right)^2=\dfrac12$.
Hence the minimum value is
$$\min\left(2,\,\frac12\right)\ =\ \frac12$$
