Question

Let $a,b,c$ be positive real numbers such that $a^3+b^3=c^3$. Prove that $a^2+b^2-c^2>6(c-a)(c-b).$

Answer 1

The given inequality may be written in the form
$7c^2-6(a+b)c-(a^2+b^2-6ab)<0$.
Putting $x=7c^2,y=-6(a+b)c,z=-(a^2+b^2-6ab)$, we have to prove that $x+y+z<0$.
Observe that $x,y,z$ are not all equal $(x>0,y<0)$.
Using the identity $x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)\left[\right(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]$,
we infer that it is sufficient to prove $x^3+y^3+z^3-3xyz<0$.
Substituting the values of $x,y,z,$ we see that this is equivalent to
$343c^6-216(a+b{)}^3{c}^3-(a^2+b^2-6ab{)}^3-126c^3(a+b)(a^2+b^2-6ab)<0$.
Using $c^3=a^3+b^3$, this reduces to $343(a^3+b^3{)}^2-216(a+b{)}^3(a^3+b^3)-(a^2+b^2-6ab{)}^3-126((a^3+b^3)(a+b)(a^2+b^2-6ab)<0$.
This may be simplified (after some tedious calculations) to,
$-a^2{b}^2(129a^2-254ab+129b^2)<0$.
But $129a^2-254ab+129b^2=129(a-b{)}^2+4ab>0$. Hence the result follows.
Remark: The best constant $\theta$ in the inequality $a^2+b^2-c^2\ge \theta (c-a)(c-b)$, where $a,b,c$
are positive reals such that $a^3+b^3=c^3$, is $\theta =2(1+2^{1/3}+2^{-1/3})$.