<s>[INST] 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
 [/INST] You can solve it also like this. By Cauchy inequality we have $$(a+b+c)(a^3+b^3+c^3)\geq (a^2+b^2+c^2)^2$$
Further: $$a^2+b^2+c^2\geq {1\over 3}(a+b+c)^2$$
so $${3\over 2}(a+b+c)^3\geq {1\over 9}(a+b+c)^4$$
and thus $$a+b+c\leq {27\over 2}$$
Since $2\mid a+b+c$ we have $a+b+c\leq 12$. 
Also, since $x^3\equiv x\pmod 3$ we have $3\mid a+b+c$ so $a+b+c\in\{0,6,12\}$.

Now if we try to find out $a,b,c$ we can assume that $a\leq b\leq c$. So if $a+b+c=12$ then $$3a^3\leq a^3+b^3+c^3 = 216\implies a\leq 4$$
Now you can find by inspection if 12 is achivable.
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