If $x^2+y^2+xy=1$ then find minimum of $x^3y+xy^3+4$

Question

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?

Answer 1

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 $2\times-1+4=2$.