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


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.

  • $\begingroup$ Can you plz explain me why $a+b+c$ is divisible by $2$ $\endgroup$ – DXT Feb 23 at 10:34
  • $\begingroup$ I did not understand that line $\endgroup$ – DXT Feb 23 at 10:36
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    $\begingroup$ The left side of starting equation is divisible by 2, so must be the right also. $\endgroup$ – Aqua Feb 23 at 10:37
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    $\begingroup$ How come you accept Rozenbergs solution? There is nothing instructive in his solution. Neither how he got this triple $(3,4,5)$ nor how hi get this formula from where he deduce $3\mid a+b+c$... $\endgroup$ – Aqua Feb 24 at 12:40
  • $\begingroup$ yours solution is great. $\endgroup$ – DXT Feb 25 at 4:30

So you have the maximum possible sum and need to restrict to integers.

What is the maximum possible sum for positive integers (hint has to be less than or equal to the sum for arbitrary reals)? Call this the target sum

Check that the target sum is even (the given condition implies that).

Then is there a solution to the equality with the target sum? If not try the next one down. You have a finite search space and cubes are quite sparse.


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