If $a,b,c>0$ and $a^3+b^3=c^3$ then prove that $$a^2+b^2-c^2>6(c-a)(c-b)$$.
I tried factoring but things got more complicated. I dont think standard AM-Gm can be applied directly. Can somebody help me to proceed? Thanks a lot.
2026-03-28 19:39:49.1774726789
If $a,b,c>0$ and $a^3+b^3=c^3$ then prove that $a^2+b^2-c^2>6(c-a)(c-b)$.
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We need to prove that $$c(a^2+b^2)-(a^3+b^3)>6c(c-a)(c-b)$$ or $$\frac{(c^3-b^3)(c-a)}{a}+\frac{(c^3-a^3)(c-b)}{b}>6c(c-a)(c-b)$$ or $$\frac{b^2+bc+c^2}{a}+\frac{a^2+ac+c^2}{b}>6c$$ which is AM-GM.
Done!