<s>[INST] If $x^2+y^2+xy=1$ then find minimum of $x^3y+xy^3+4$ 
If $x,y \in \mathbb{R}$ and $x^2+y^2+xy=1$ then find the minimum value of $x^3y+xy^3+4$

My Attempt:
$x^3y+xy^3+4$
$\Rightarrow xy(x^2+y^2)+4$
$\Rightarrow xy(1-xy)+4$ (from first equation)
$\Rightarrow xy-(xy)^2+4  =f(x)$
For minimum value, $\frac{df(x)}{dx}=0$.
$\Rightarrow \frac{df(x)}{dx}=(y-2xy²) + \frac{dy}{dx}(x-2yx²)=0$
How should I proceed from here?
 [/INST] By AM-GM, $x^2+y^2=|x|^2+|y|^2\ge2|xy|$ so $r^2=x^2+y^2$ has extrema $\tfrac23,\,2$ respectively achieved by$$2xy=x^2+y^2\implies1=(1+1/2)(x^2+y^2)$$and$$-2xy=x^2+y^2\implies1=(1-1/2)(x^2+y^2).$$Note the minimum of $r^2(1-r^2)$ on $[\tfrac23,\,2]$ occurs at $r^2=2$. In particular, $x=-y=1$ minimizes $r^2(1-r^2)+4$ as $2\times-1+4=2$.
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