Prove that $\sqrt[3]{\frac{a^3+b^3+c^3+3abc}{2}} \geq \max{\{a,b,c\}}$ 
Let $a,b,$ and $c$ be the side lengths of a triangle. Prove that $$\sqrt[3]{\dfrac{a^3+b^3+c^3+3abc}{2}} \geq \max{\{a,b,c\}}.$$

The $a^3+b^3+c^3+3abc$ makes me think of $(a+b+c)^3$, and I don't think a ravi substitution will work here. I think using the triangle inequality also might help but it is going to be hard to simplify that.

Like vadim123 said, we only have to show that with $a+b\geq 1$, $a\leq 1,b\leq 1$ we have
$$a^3+b^3+3ab\geq 1$$
That's because of
$$a^3+b^3+3ab\geq a^3+(1-a)^3+3a(1-a)=3a^2-6a+4\geq 1$$
You see the last inequality by calculating the minimum.
