Prove $a^3+b^3+3abc>c^3$ where a,b,c are triangle sides If $a,b,c$ are triangle sides prove
$a^3+b^3+3abc>c^3$
 A: 
Triangle inequality: If $~a,~ b, ~$and$~ c~$ are the lengths of the sides of the triangle, with no side being greater than $~c~$, then the triangle inequality states that
  $~c\leq a+b~$.

Now to prove the given inequality $~a^3+b^3+3abc>c^3~$, we have to use the above property. So we have
$$a^3 + b^3 + 3abc = (a + b)(a^2 - ab + b^2) + 3abc $$
$$~~~~~~~~~~~~~~~> c(a^2 - ab + b^2) + 3abc $$
$$~~~~~~~~~~~~~~~~~~~~~~= c[a^2 -ab+ b^2+3ab]$$
$$~~= c(a + b)^2 $$
$$\implies a^3 + b^3 + 3abc > c^3~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
A: 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 $F(c)<F(a+b)=0.$
