Let $x:[0,\infty[\to[0,\infty[$ be a differentiable function satisfying
$$\dot x(t)=\frac{c}{1-\varepsilon} x(t)^\varepsilon$$ for all $t\in[0,\infty[$, a fixed $c\ge 0$ and a fixed $\varepsilon\in[0,1[$. Furthermore, assume that $x(0)=0$.
I want to prove that $x(t)\le (ct)^{1/(1-\varepsilon)}$ for all $t\ge 0$.
Note: Morally speaking, there are only the solutions $x=0$ and $x(t)=(ct)^{1/(1-\varepsilon)}$. The latter solution is obtained by separating variables, i.e. by writing $$\dot x(t)x(t)^{-\varepsilon}=\frac{c}{1-\varepsilon}$$ and then by integrating. However, that approach requires $x(t)\neq 0$, which I can't assume in my case.
Note 2: The usual uniqueness result for ODEs does not hold in this case since $x\mapsto x^\varepsilon$ is not Lipschitz-continuous around $x=0$.
Let $x(t_0)> 0$ for some $t_0 >0$ (if $x(t)$ is identically zero you are done). Let $I= (a, b)$, where $a\ge 0$ and $b\le +\infty$, be the maximum interval in $[0, \infty)$ containing $t_0$ and that $x(t)> 0$ in $I$. Then on $I$ you can divide by $x^\epsilon$ and use $x(a) = 0$ to get
$$ x(t)^{1-\epsilon} = c(t-a)\Rightarrow x(t) = (c(t-a))^{1/(1-\epsilon)}. $$
In particular, $ b = +\infty$ and $x(t) = 0$ in $[0, a]$ (if not, then $x(t)>0$ for some $a$ in this interval, the above argument implies that $x(a)>0$). Thus if $x(t)$ is not identically zero, $x(t)$ is of the form
$$x_a(t) = \begin{cases} 0 & \text{ if } x\le a, \\ (c(t-a))^{1/(1-\epsilon)} & \text{ if } t > a.\end{cases}$$
and your inequality is satisfied.