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:
$x^2+y^2-z^2&gt;6(z-x) (z-y)$ i.e., $\underbrace{(z^2+zx+x^2) (z^2+zy+y^2) (x^2+y^2-z^2)}_{=E\text{ (say)}}&gt;6(z^3-x^3) (z^3-y^3)=6x^3y^3.$
(Here one thing which I noticed is that $(x^2+y^2-z^2)&gt;0,$ since each of the terms on the LHS except this one is positive and $6x^3y^3$ is also positive for $x, y, z&gt;0.$)
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 $E&gt;6x^3y^3.$ 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.
Th proof by Bjkjdz. The inequality equivalent to
$${x^2}z + {y^2}z - {z^3} &gt; 6z(z - x)(z - y),$$
or
$${x^2}(z - x) + {y^2}(z - y) &gt; 6z(z - x)(z - y),$$
or
$$\dfrac{{{x^2}}}{{z - y}} + \dfrac{{{y^2}}}{{z - x}} &gt; 6z.$$
But
$$\left\{ \begin{array}{l} {x^2} = \dfrac{{{x^3}}}{x} = \dfrac{{{z^3} - {y^3}}}{x} = \dfrac{{(z - y)({z^2} + yz + {y^2})}}{x} \\ {y^2} = \dfrac{{{y^3}}}{y} = \dfrac{{{z^3} - {x^3}}}{y} = \dfrac{{(z - x)({z^2} + zx + {x^2})}}{y} \\ \end{array} \right.$$
Thefore, we will show that
$$\dfrac{{{y^2} + yz+ {z^2}}}{x} + \dfrac{{{z^2} + zx + {x^2}}}{y} &gt; 6z,$$
equivalent to
$$\left( {{x^3} + y{z^2} + {y^2}z} \right) + \left( {{y^3} + {x^2}z + x{z^2}} \right) &gt; 6xyz.$$
Which is true by the AM-GM inequality.
Hint: eliminate $z$ from the system
$$ \eqalign{x^3 + y^3 &amp;= z^3 \cr 
      x^2 + y^2 - z^2 &amp;= 6 (z-x)(z-y)\cr} $$
and show the only real solutions of the resulting equation are $x=0$ and $y=0$.