Question

If $x,y,z$ are non-negative integers such that $2(x^3+y^3+z^3)=3(x+y+z{)}^2$, then maximum value of $x+y+z$ is

Answer 1

Given : $2(x^3+y^3+z^3)=3(x+y+z{)}^2$
Let $u=f\left(t\right)=t^3$


Let three points on the graph be $A(x,x^3),B(y,y^3),C(z,z^3)$

We know that, the centroid $G$ lies inside $△ABC$, so
$GL\ge ML\Rightarrow \frac{x^3+y^3+z^3}{3}\ge f\left(\frac{x+y+z}{3}\right)\Rightarrow \frac{x^3+y^3+z^3}{3}\ge \frac{(x+y+z{)}^3}{27}\Rightarrow \frac{(x+y+z{)}^2}{2}\ge \frac{(x+y+z{)}^3}{27}\Rightarrow (x+y+z)\le \frac{27}{2}$

As $x,y,z$ are integers, so
When $x+y+z=13$, then
$x^3+y^3+z^3=\frac{3\times 169}{2}$
This is not possible.

When $x+y+z=12$, then
$x^3+y^3+z^2=\frac{3\times 144}{2}$
This is possible.

Hence, the maximum possible value of $x+y+z$ is $12$.