Prove $a^2+b^2-c^2>6(c-a)(c-b) for $a^3+b^3=c^3$

[anemone]

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

 

[jvicens]

Thank you for bringing up this interesting problem. I am always excited to explore new mathematical concepts and solve challenging equations.

To prove the given inequality, let us start by simplifying the expression $a^2+b^2-c^2$. We can rewrite it as $(a+b)^2-2c^2$. Similarly, we can express $6(c-a)(c-b)$ as $6c^2-6ac-6bc+6ab$. Now, we can substitute $a^3+b^3$ for $c^3$ in the given equation, giving us $a^2+b^2=c^2-a^3-b^3$.

Substituting these expressions into the original inequality, we get: $(a+b)^2-2c^2 > 6c^2-6ac-6bc+6ab$. Rearranging the terms, we have $(a+b)^2+4c^2 > 6ac+6bc+6ab$. Using the Cauchy-Schwarz inequality, we can rewrite the right side of the inequality as $6(a+b)(c+a)(c+b)$. Therefore, we have $(a+b)^2+4c^2 > 6(a+b)(c+a)(c+b)$.

Now, let us focus on the left side of the inequality. Using the AM-GM inequality, we know that $(a+b)^2 \geq 4ab$. Substituting this into the inequality, we get $4ab+4c^2 > 6(a+b)(c+a)(c+b)$. Using the AM-GM inequality again, we know that $4ab \geq 12c(c-a)(c-b)$. Substituting this into the inequality, we have $12c(c-a)(c-b)+4c^2 > 6(a+b)(c+a)(c+b)$.

Finally, we can factor out $6c$ from the right side of the inequality, giving us $12c(c-a)(c-b)+4c^2 > 6c(a+b)(c+a)(c+b)$. Using the given equation, we know that $a^3+b^3=c^3$, which can be rewritten as $c^3-a^3-b^3=0$. Factoring this, we have $(c-a)(c-b)(