Suppose we have a $1$-dimensional differential inequality
$$\frac{dx}{dt} \leq x - x^3 $$
We can apply the Comparison principle to claim that if $y(t)$ is the solution to $\frac{dy}{dt} = y - y^3$, then $x(t) \leq y(t)$ (assuming $x(0) \leq y(0)$. Can we extend similar argument to a $2$-dimensional system? For example, let us consider the following system of equations
$$\frac{dx_1}{dt}=x_1-x_2, \quad \quad \quad \frac{dx_2}{dt} \leq x_1 + x_2 - \frac{x_2^4}{x_1^2}$$
Is the solution to following
$$\frac{dy_1}{dt}=y_1-y_2, \quad \quad \quad \frac{dy_2}{dt} = y_1 + y_2 - \frac{y_2^4}{y_1^2}$$
related with the solution of the original problem. Specifically, can we apply the Comparison principle to first say that $x_2(t) \leq y_2 (t)$ and then subsequently use it to claim $x_1(t) \geq y_1(t)$?
I don't know about the specific example you give. However, the answer is "no" for general problems, even if they are linear. Consider the structure:
\begin{align} x' &= f(x,y)\\ y' &=g(x,y) \leq h(x,y) \end{align} Specifically, compare the following two systems:
System 1
\begin{align} x' &= x-y\\ y' &= \underbrace{2x-y-1}_{g(x,y)} \end{align}
System 2
\begin{align} w' &= w-z\\ z' &= \underbrace{2w-z}_{h(w,z)} \end{align} Suppose initial conditions are $(x(0),y(0))=(w(0),z(0))=(0,0)$. Then $(w(t),z(t))=(0,0)$ for all $t \geq 0$. However, $(x(t),y(t))$ has solution: \begin{align} x(t) &= 1-\cos(t)\\ y(t) &= 1-\sin(t) - \cos(t) \end{align} and indeed $y(t)$ takes both positive and negative values over $t \in [0, \infty)$. So we cannot say that $y(t) \leq \underbrace{z(t)}_{0}$ for all $t$.