Find all positive integers $x, y, z$ such that $x^2y+y^2z+z^2x = 3xyz$.
I first tried to solve it by spilting $xyz$ to every expression and factor it. But it fails.
I notice that it is true for $\left(x,y,z\right) = \left(0,0,0\right)$. But how can I prove all of then and find the values ?
On
If you assume one of the variables is $0$ (say $x=0$), then we must have $y^{2}z = 0$, so either $y=0$, $z=0$ or both, hence the triples $\left(0,0,k\right)$, $\left(0,k,0\right)$ and $\left(k,0,0\right)$ are solutions for $k \in \mathbb{Z}$, now, if $xyz \neq 0$ we can divide both sides of the equation by $xyz$ and we get
$$ \frac{x}{z} +\frac{y}{x}+\frac{z}{y} = 3. $$
By $AM \geq GM$ we have that $\frac{x}{z} +\frac{y}{x}+\frac{z}{y}\geq 3$ with equality iff
$$ \frac{x}{z} =\frac{y}{x}=\frac{z}{y} $$
which can only happens if $x=y=z$, so the only other solutions are $\left(k,k,k\right)$ with $k\in \mathbb{Z}$.
By AM-GM inequality, you have:
$$x^2y+y^2z+z^2x=3xyz≥3xyz$$
Equality holds iff, when $x^2y=y^2z=z^2x\thinspace.$
Thus we have:
$$\begin{align}&\begin{cases}x^2=yz\\y^2=xz\\z^2=xy\end{cases}\\ \implies &\left(\frac xy\right)^2=\frac yx\\ \implies &\left(\frac xy\right)^3=1\\ \implies &x=y\end{align}$$
This yields, $x=y=z$ are only possible solution. Note that, $(x,y,z)=(x,x,x)\in\mathbb R^3_{≥0}$ are also solution set.