Maximum value of expression $a+b+c$ 
If $a,b,c$ are non negative integers such that  $$2(a^3+b^3+c^3)=3(a+b+c)^2.$$
Then maximum value of $a+b+c$ is ?

My Try: Using Jensen Inequality
Let $f(x)=x^3$. Then $f''(x)>0$ for $x>0$ is convex function
So $$\frac{f(a)+f(b)+f(c)}{3}\geq f\bigg(\frac{a+b+c}{3}\bigg)$$
$$\frac{a^3+b^3+c^3}{3}\geq \bigg(\frac{a+b+c}{3}\bigg)^3\cdots (1)$$
From given condition $$\frac{a^3+b^3+c^3}{3}=\frac{(a+b+c)^2}{2}\cdots (2)$$
So we have $$\frac{(a+b+c)^2}{2}\geq \frac{(a+b+c)^3}{27}$$
$$a+b+c\leq \frac{27}{2}=13.5$$
equality hold when $a=b=c=4.5$
but $a,b,c$ are non negative integers
Could some help me to solve it, Thanks

By your work $$a+b+c\leq13,$$ but since
$$3(a+b+c)^2=2(a^3+b^3+c^3)=2(a^3+b^3+x^3-3abc+3abc)=$$
$$=2(a+b+c)(a^2+b^2+c^2-ab-ac-bc)+6abc,$$ we obtain that $a+b+c$ is divisible by $3$.
Thus, we see that $$a+b+c\leq12.$$ But $$(a,b,c)=(3,4,5)$$ is valid, which says that $12$ is a maximal value.
Actually, your inequality we can get also by Holder:
$$a^3+b^3+c^3=\frac{1}{9}(1+1+1)^2(a^3+b^3+c^3)\geq\frac{1}{9}(a+b+c)^3.$$
