<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] $$x^3+3xy+y^3-1=(x+y-1)(x^2+y^2-xy+x+y+1)=0$$
So either $y=1-x$ or $(x,y)=(-1,-1)$
Subbing this in gives $x^2+(1-x)^2$ or $2$ to be minimized. Considering the first simple calculus gives us that $\frac{dy}{dx}=2x-2(1-x)=4x-2$. Setting this equal to zero gives $x=y=\frac{1}{2}$ as a possible minimum/maximum. Looking at the second derivative $\frac{d^2y}{dx^2}=4$ shows it is a minimum. Evaluating it gives $x^2+y^2=\frac{1}{2}$ which is less than the value at $(x,y)=(-1,-1)$ so the minimum value is $\frac{1}{2}$ at $(x,y)=\left(\frac{1}{2},\frac{1}{2}\right)$.
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