<s>[INST] Prove $a^3+b^3+3abc>c^3$ where a,b,c are triangle sides If $a,b,c$ are triangle sides prove
$a^3+b^3+3abc>c^3$
 [/INST] 
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
$$a^3 + b^3 + 3abc = (a + b)(a^2 - ab + b^2) + 3abc $$
$$~~~~~~~~~~~~~~~> c(a^2 - ab + b^2) + 3abc $$
$$~~~~~~~~~~~~~~~~~~~~~~= c[a^2 -ab+ b^2+3ab]$$
$$~~= c(a + b)^2 $$
$$\implies a^3 + b^3 + 3abc > c^3~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
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