If $x, y, z\in\mathbb R^+ $ and $x^3+y^3=z^3,$ then prove that $x^2+y^2-z^2>6(z-x) (z-y). $ I made several unsuccessful attempts. Still couldn't think of a proper way to prove the inequality. Please suggest how to approach this problem. Thanks in advance.
EDIT 1. My approach (that I was talking about):
Given: $z^3=x^3+y^3.$
We have to prove:
$x^2+y^2-z^2>6(z-x) (z-y)$ i.e., $\underbrace{(z^2+zx+x^2) (z^2+zy+y^2) (x^2+y^2-z^2)}_{=E\text{ (say)}}>6(z^3-x^3) (z^3-y^3)=6x^3y^3.$
(Here one thing which I noticed is that $(x^2+y^2-z^2)>0,$ since each of the terms on the LHS except this one is positive and $6x^3y^3$ is also positive for $x, y, z>0.$)
Using AM $\ge$ GM, we have:
$E\ge 3zx\cdot3zy\cdot(x^2+y^2-z^2)\ge 9xyz^2(2xy-z^2).$
From here I couldn't think of a proper way to prove $E>6x^3y^3.$ But I'm still working on it. At present I'm trying to manipulate the expression $9xyz^2(2xy-z^2)$ to get the job done. If I find something useful I'll update here.

Let $x^2+y^2=2uxy$.
Thus, $u\geq1$ and we need to prove that:
$$x^2+y^2-\sqrt[3]{(x^3+y^3)^2}>6\left(\sqrt[3]{x^3+y^3}-x\right)\left(\sqrt[3]{x^3+y^3}-y\right)$$ or
$$x^2+y^2-6xy+6(x+y)\sqrt[3]{x^3+y^3}-7\sqrt[3]{(x^3+y^3)^2}>0$$ or
$$(x^2+y^2-6xy)^3+216(x+y)^3(x^3+y^3)-343(x^3+y^3)^2+$$
$$+126(x^2+y^2-6xy)(x+y)(x^3+y^3)>0$$ or
$$4(u-3)^3+432(u+1)^2(2u-1)-343(u+1)(2u-1)^2+$$
$$+252(u-3)(u+1)(2u-1)>0$$ or
$$129u-127>0,$$ which is obvious.
