Maximum value of expression $a+b+c$

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: $$a^3+b^3+c^3=\frac{1}{9}(1+1+1)^2(a^3+b^3+c^3)\geq\frac{1}{9}(a+b+c)^3.$$

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 $a+b+c\in\{0,6,12\}$.

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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.