<s>[INST] What is the minimum value of $x^2+y^2$ subject to $x^3+y^3+xy=1$? 
What is the minimum value of $x^2+y^2$ under the constraint $x^3+y^3+xy=1$? Please do not use partial differentials (multivariable calculus) or Lagrange multipliers. You can use elementary algebra or single variable calculus.

I plotted the graph of $x^3+y^3+xy=1$. It seems the minima occurs when $x=y$ but I don't know why that will be true. Any ideas?
 [/INST] The cubic $x^3+y^3+xy+1$ is symmetric in $y=x$. So lets rotate it by $\frac{\pi}{4}$. Let $x=\frac{x`}{\sqrt{2}}+\frac{y`}{\sqrt{2}}$ and $y=\frac{x`}{\sqrt{2}}-\frac{y`}{\sqrt{2}}$. Note that the function to minimize becomes $x`^2+y`^2$.
$$\left(\frac{x`}{\sqrt{2}}+\frac{y`}{\sqrt{2}}\right)^3+\left(\frac{x`}{\sqrt{2}}-\frac{y`}{\sqrt{2}}\right)^3+\left(\frac{x`}{\sqrt{2}}+\frac{y`}{\sqrt{2}}\right)\left(\frac{x`}{\sqrt{2}}-\frac{y`}{\sqrt{2}}\right)-1=0$$
$$\frac{3x`y`^2}{\sqrt{2}}+\frac{x`^3}{\sqrt{2}}-\frac{y`^2}{2}+\frac{x`^2}{2}-1=0$$
We can solve for $y`^2$.
$$y`^2=\frac{-\sqrt{2}x`^3-x`^2+2}{3\sqrt{2}x`-1}$$
As this is positive solutions will only exist for $-\sqrt{2}x`^3-x`^2+2\ge0$. Note the root of $-\sqrt{2}x`^3-x`^2+2=0$ can be found by solving: $-2\left(2\left(\frac{x`}{\sqrt{2}}\right)^3+\left(\frac{x`}{\sqrt{2}}\right)^2-1\right)=0$. Also note that $y`^2$ is strictly decreasing on this domain.
Using calculus to find the minimum of $x`^2+y`^2$ gives a value outside this domain so the minimum must be at the end of the domain, i.e. at the real root of $2\left(\frac{x`}{\sqrt{2}}\right)^3+\left(\frac{x`}{\sqrt{2}}\right)^2-1=0$. And the corresponding $y`$ is $0$. So rotating back to $x$ and $y$ shows that the minimum is on the line $x=y$ and is the real root of $2x^3+x^2-1$.
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