A question about following ODE. I was reading "Ricci flow, an introduction" by B. Chow and D. Knopf and I was stuck by the equation on Page 13. As below, $x,y,z>0$ satisfying
\begin{equation}
\begin{array}{ll}
\dfrac{\mathrm{d}x}{\mathrm{d}t} \!\!\!\!\!&=-8+4\dfrac{y^2+z^2-x^2}{yz},\\
\dfrac{\mathrm{d}y}{\mathrm{d}t} \!\!\!\!\!&=-8+4\dfrac{x^2+z^2-y^2}{xz},\\
\dfrac{\mathrm{d}z}{\mathrm{d}t} \!\!\!\!\!&=-8+4\dfrac{x^2+y^2-z^2}{xy}.
\end{array}
\end{equation}
We assume that $x(0)\geqslant y(0)\geqslant z(0)$, then we can calculate that
\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}t}(x-z)=4(x-z)\frac{y^2-(x+z)^2}{xyz}.
\tag{$\ast$}
\end{equation}
Then the author conclude that $x(t)\geqslant y(t)\geqslant z(t)$  persist for as long as the solution exists.
Here is my question: how can we judge $x(t)\geqslant y(t)\geqslant z(t)$ from  $(\ast)$? My attempt is
\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}t}(\mathrm{ln}|x-z|)=4\frac{y^2-(x+z)^2}{xyz}.
\end{equation}
But then what should I do?
 A: From the equation
\begin{eqnarray}
\frac{\mathrm{d}}{\mathrm{d}t}(x-z)=4(x-z)\frac{y^2-(x+z)^2}{xyz}
\end{eqnarray}
it follows that $(x-z)(t)$ has an equilibrium at $0$. Hence a solution $(x-z)(t)$ can never change sign. In particular, this means that if $x(0) \ge z(0)$ then also $x(t) \ge z(t)$ for all $t \ge 0$.
Likewise, you can show $x(t) \ge y(t)$ and $y(t) \ge z(t)$ for all $t\ge 0$ by a similar argument.
A: If $\varphi=4\dfrac{y^2-(x+z)^2}{xyz}$ then
\begin{eqnarray}
\frac{\mathrm{d}}{\mathrm{d}t}(x-z)=4(x-z)\frac{y^2-(x+z)^2}{xyz}
\qquad\qquad (\star)
\end{eqnarray}
can be rewritten as
\begin{eqnarray}
\frac{\mathrm{d}}{\mathrm{d}t}(x-z)=\varphi(t)(x-z)
\end{eqnarray}
which implies that
$$
\exp\Big(-\int_0^t\varphi(s)\,ds\Big)\frac{\mathrm{d}}{\mathrm{d}t}(x-z)
-\exp\Big(-\int_0^t\varphi(s)\,ds\Big)\varphi(t)(x-z)=0
$$
and hence,
$$
\Big((x-z)\exp\Big(-\int_0^t\varphi(s)\Big)'=0
$$
and thus, for some constant $c\in\mathbb R$,
$$
(x-z)\exp\Big(-\int_0^t\varphi(s)\Big)=c
\quad\Longrightarrow\quad x(t)-z(t)=c\exp\Big(-\int_0^t\varphi(s)\Big).
$$
Thus $x(t)-z(t)$ maintains sign.
Similarly, we can show that $x-y$, $y-z$ also maintain sign by obtaining analogous equation for $(x-y)'$ and $(y-z)'$ as in $(\star)$ for $(x-z)'$.
