Prove $a^3+b^3+3abc>c^3$ where a,b,c are triangle sides If $a,b,c$ are triangle sides prove
$a^3+b^3+3abc>c^3$

Method 1. For any numbers $a,b,c$ let $F(c)=c^3-3abc-(a^3+b^3).$ Observe that $F(a+b)=0$ so $c-(a+b)$ is a factor of the polynomial $F(c).$ Using synthetic division we obtain $$F(c)=[c-(a+b)]\cdot [c^2+c(a+b)+(a^2-ab+b^2)].$$ Now if $a,b,c$ are positive and $c<a+b$ then $$c-(a+b)<0$$ while $$c^2+c(a+b)+(a^2-ab+b^2)>(a^2-ab+b^2)=(a-b)^2+ab>0,$$ so $F(c)<0.$
Method 2. The inequality is obvious if $c\le \max (a,b).$ With $F(c)$ as in Method 1., with positive $a,b:$ We have $F'(c)=3(c^2-ab)>0$ whenever    $c>\max(a,b),$ so $F(c)$ is strictly increasing for $c>\max (a,b).$ So if  $a+b>c>\max(a,b)$ then $F(c)<F(a+b)=0.$
