<s>[INST] 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.
 [/INST] The solution is due to @Calvin Lin.
Problem: Find the best constant $C$
such that $x^2+y^2 - 1 \ge C(1-x)(1-y)$ holds for all
$x, y \ge 0$ with $x^3+y^3 = 1$.
Solution: The best constant is $C = 2^{4/3} + 2^{2/3} + 2$. Let us prove it.
Let $w = x + y$. Since $x^3+y^3 \le (x+y)^3 \le 4(x^3 + y^3)$, we have $1\le w \le \sqrt[3]{4}$.
We have
\begin{align}
&(x+y)[x^2+y^2 - 1 - C(1-x)(1-y)]\\
=\ & x^3+y^3 - (C-1)xy(x+y) + C(x+y)^2 - (C+1)(x+y)\\
=\ & x^3+y^3 - \frac{C-1}{3}[(x+y)^3-x^3-y^3] + C(x+y)^2 - (C+1)(x+y)\\
=\ & 1 - \frac{C-1}{3}[(x+y)^3-1] + C(x+y)^2 - (C+1)(x+y)\\
=\ & \frac{1}{3}(w-1)^2[2 + w - C(w-1)].
\end{align}
From $2 + w - C(w-1) \ge 0$ for $1\le w \le \sqrt[3]{4}$,
we have
$$C \le \inf_{1 < w \le \sqrt[3]{4}} \frac{2+w}{w-1} = 2^{4/3} + 2^{2/3} + 2.$$
On the other hand, when $C = 2^{4/3} + 2^{2/3} + 2$, we have
$$2 + w - C(w-1) = (2^{4/3} + 2^{2/3} + 1)(\sqrt[3]{4} - w) \ge 0$$
for $1\le w \le \sqrt[3]{4}$. We are done.
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