Question

Solve the system of equations for positive real numbers $\frac{1}{xy}=\frac{x}{z}+1,\frac{1}{y^z}=\frac{y}{x}+1,\frac{1}{zx}=\frac{z}{y}+1$.

Answer 1

We can write given equation as,
$z=x^{2}y+xyz,x=y^z+xyz,y=z^{2}x+xyz$.
$\therefore z-x^{2}y=x-y^{2}z=y-z^{2}x$
If $x=y$, then $y^{2}z=z^{2}x$ and hence $x^{2}z=z^{2}x$.
This implies that $z=x=y$. Similarly, $x=z$ implies that $x=z=y$.
Hence if any two of x, y, z are equal, then all are equal. Suppose no two of x, y, z are equal.
We may take x is the largest among x, y, z so that x > y and x > z. Here we have two possibilities: y > z and z > y.
Suppose x > y > z. Now $z-x^{2}y=x-y^{2}z=y-z^{2}x$ shows that
$y^{2}z>z^{2}x>x^{2}y$
But $y^{2}z>z^{2}x$ and $z^{2}x>x^{2}y$ give $y^2>xy$. Hence $\left(y^2\right)\left(z^2\right)>\left(zx\right)\left(xy\right)$.
This gives $yz>x^2$. Thus $x^3<xyz=\left(xz\right)y<\left(y^2\right)y=y^3$. This forces x < y contradicting x > y.
Similarly, we arrive at a contradiction if $x>z>y$. The only possibility is $x=y=z$.
For $x=y=z$, we get only one equation $x^2=1/2$. Since x > 0, $x=1/\sqrt{2}=y=z$.