Prove that $a^3 + b^3 + 3abc > c^3$ Suppose $a,b,c$ are the sides of a non-degenerate triangle. Prove that $a^3 + b^3 + 3abc > c^3.$

I was thinking that this inequality looked suspiciously like $a^3 + b^3 + c^3 - 3abc,$ which factors as $(a + b + c) (a^2 - a b + b^2 - a c - b c + c^2).$ However, I have little to no idea how to get that. Can someone give me a hint?

By your hint we need to prove that: $$(a+b-c)(a^2+b^2+c^2-ab+ac+bc)>0.$$
Can you end it now?
$$a^2+b^2+c^2-ab+ac+bc=b(b+c-a)+a^2+c^2+ac>0.$$
