<s>[INST] 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.
 [/INST] Let $x=u+v$ and $y=u-v$.  Then $x^2+y^2=2(u^2+v^2)$, while the equation $x^3+3xy+y^3=1$ becomes $2u^3+6uv^2+3(u^2-v^2)=1$, or
$$2u^3+3u^2-1=-3v^2(2u-1)$$
Factoring the cubic on the left hand side, we obtain
$$(u+1)^2(2u-1)=-3v^2(2u-1)$$
So either $2u-1=0$, in which case $v$ can be anything, or else $(u+1)^2=-3v^2$, which, because of the minus sign, is possible only if $u+1=v=0$.  The latter possibility leads to 
$$x^2+y^2=2((-1)^2+0^2)=2$$ 
while the former leads to 
$$x^2+y^2=2\left(\left(1\over2\right)^2+v^2\right)={1\over2}+2v^2$$
which is clearly minimized when $v=0$.  So the minimum value of $x^2+y^2$ is $1\over2$, and it occurs at $(x,y)=({1\over2}+0,{1\over2}-0)=({1\over2},{1\over2})$.
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