# 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.

• Without loss, assume $c$ is maximal, and rescale so that $c=1$. Now $a+b\ge 1$, $a,b\le 1$, and we want $\sqrt[3]{(a^3+b^3+3ab+1)/2}\ge 1$. – vadim123 Jan 6 '16 at 15:58
• Ravi substitution works fine . – user252450 Jan 6 '16 at 16:01
• @ComplexPhi How so? – user19405892 Jan 6 '16 at 16:05
• After the usual substitutions : $a=x+y$ , $b=y+z$ , $c=z+x$ and assuming the maximum is $a$ the inequality is : $$2z(3x^2+3y^2+z^2+3xy+3yz+3xz) \geq 0$$ – user252450 Jan 6 '16 at 16:11

Make the usual Ravi substitutions : $a=x+y$ , $b=y+z$ , $c=x+z$ and assume that $\max(a,b,c)=a$ .
$$2z(3x^2+3y^2+z^2+3xy+3yx+3zx) \geq 0$$ which is obvious .
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.