$x^2 + y^2+xy = 1$ , then find the minimum of $x^3 y + xy^3 +4$ x and y belongs to real numbers. $ x^2 + y^2+xy = 1 $.  then find the minimum value of $x^3 y + xy^3 +4$.
I assume  $ x = r \sin (w)$ and $ y = r\cos(w) $. $  x^3 y + xy^3 +4 = L $  which give me $ \frac{2}{3} \le r^2 \le 2 $ I am stuck after that.Its my  Humble request to help me after that.

For $x=1$ and $y=-1$ we'll get a value $2$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$x^3y+y^3x+4\geq2$$ or
$$x^3y+y^3x+2(x^2+xy+y^2)^2\geq0$$ or
$$2x^4+5x^3y+6x^2y^2+5xy^3+2y^4\geq0$$ or
$$2x^4+4x^3y+2x^2y^2+x^3y+2x^2y^2+xy^3+2x^2y^2+4xy^3+2y^4\geq0$$ or
$$(x+y)^2(2x^2+xy+2y^2)\geq0.$$
Done!
