An inequality involving real numbers Let $x,y,z$ be real numbers such that $xyz=-1$. Prove that
$$\sqrt[3/2]{\frac{3}{2}}\geq E:=\frac{4(x^3+y^3+z^3)}{(x^2+y^2+z^2)^2}$$
I tried to imitate an idea by River Li but it does not work. The idea is to find a function $f$ such that for all $x,y>0$,
$$E\leq f(x+y)$$
and then use calculus to show $\displaystyle f_{\max}=\sqrt[3/2]{\frac{3}{2}}$.
For instance, $$x^2+y^2\geq\frac{(x+y)^2}{2},\qquad z^2=\frac{1}{x^2y^2}\geq\frac{16}{(x+y)^4},\qquad z^3=-\frac{1}{x^3y^3}\leq-\frac{64}{(x+y)^6}$$
Unfortunately, there does not exist any function $g$ such that $x^3+y^3\leq g(x+y)$.
 A: Let $x+y+z=3u$, $xy+xz+yz=3v^2,$ where $v^2$ can be negative, and $xyz=w^3$.
Thus, we need to prove that:
$$\left(\frac{3}{2}\right)^{\frac{2}{3}}(x^2+y^2+z^2)^2\geq-4\sqrt[3]{xyz}(x^3+y^3+z^3).$$
Now, we can assumme that $x^3+y^3+z^3>0,$   otherwise   the inequality is obviously true.
But, if so $$x^3+y^3+z^3-3xyz>0$$ or
$$(x+y+z)\sum_{cyc}(x-y)^2>0,$$ which gives $u>0$.
Id est, we need to prove that:
$$3\left(\frac{3}{2}\right)^{\frac{2}{3}}(3u^2-2v^2)^2\geq-4w(9u^3-9uv^2+w^3)$$ or
$$12\left(\frac{3}{2}\right)^{\frac{2}{3}}v^4-\left(36uw+36\left(\frac{3}{2}\right)^{\frac{2}{3}}u^2\right)v^2+27\left(\frac{3}{2}\right)^{\frac{2}{3}}u^4+36u^3w+4w^4\geq0,$$ for which it's enough to prove that
$$324u^2\left(w+\left(\frac{3}{2}\right)^{\frac{2}{3}}u\right)^2-12\left(\frac{3}{2}\right)^{\frac{2}{3}}\left(27\left(\frac{3}{2}\right)^{\frac{2}{3}}u^4+36u^3w+4w^4\right)\leq0$$ or
$$18\left(\frac{3}{2}\right)^{\frac{2}{3}}u^3w+27u^2w^2-4\left(\frac{3}{2}\right)^{\frac{2}{3}}w^4\leq0$$ or
$$18\left(\frac{3}{2}\right)^{\frac{2}{3}}u^3+27u^2w-4\left(\frac{3}{2}\right)^{\frac{2}{3}}w^3\geq0,$$ which is true by AM-GM:
$$18\left(\frac{3}{2}\right)^{\frac{2}{3}}u^3-4\left(\frac{3}{2}\right)^{\frac{2}{3}}w^3+27u^2w\geq$$
$$\geq3\sqrt[3]{\left(9\left(\frac{3}{2}\right)^{\frac{2}{3}}u^3\right)^2\cdot\left(-4\left(\frac{3}{2}\right)^{\frac{2}{3}}w^3\right)}+27u^2w=0$$ and we are done!
