A polynomial differential inequality I have come across the following differential inequality$$
y'(t)\leq C_1y(t)^\theta-C_2y(t)^\beta
$$
where $C_1,C_2,\theta,\beta\geq 0$. My goal is to estimate $y(t)$. The case $C_1=0$ and $C_2\neq 0$ was dealt with a Gronwall-like inequality and I got an estimate in the form of a constant times a power of $t$.
I am wondering if there is a comparison principle so that I can estimate $y(t)$ in the case of strictly positive constants. Any ideas?
Edit: In the meantime, I tried to multiply both sides by $y^{-\theta}$ to get$$
\frac{d}{dt}\left(\frac{y^{-\theta+1}}{\theta+1}  \right)\leq C_1-C_2y^{\beta-\theta}
$$but this doesn't seem to lead anywhere, even changing functions, since I still have $C_1>0$.
 A: If $\theta, \beta \not\in\mathbb N \,\cup \left\{\frac{p}q\,;\,p,q\in\mathbb N \,\text{ are odd}\right\}$ and $y<0$, the RHS of the inequality is not well defined.
I will hence assume, that $y\geq0$.

This is not quite a full answer since there are additional supposition, but still.
We try to apply the integral form of Gronwall's inequality.
Proceeding from your work, rewrite your inequality as
\begin{align}
&\frac{y(t)^{1-\theta}}{1-\theta}-\frac{y(0)^{1-\theta}}{1-\theta}\leq C_1t-\int_0^t\,C_2\,(1-\theta)\,y(x)^{\beta-1}\cdot\frac{y(x)^{1-\theta}}{1-\theta}\,dx\\
\iff \quad&
\frac{y(t)^{1-\theta}}{1-\theta} \leq
\left(\frac{y(0)^{1-\theta}}{1-\theta}+C_1t\right) +
\int_0^t\,C_2\,(\theta-1)\,y(x)^{\beta-1}\cdot\frac{y(x)^{1-\theta}}{1-\theta}\,dx.
\end{align}
Then, with $u(t)=\frac{y(t)^{1-\theta}}{1-\theta}$, $a(t)=\frac{y(0)^{1-\theta}}{1-\theta}+C_1t$ and $b(t)=C_2\,(\theta-1)\,y(t)^{\beta-1}$, the inequality can be written as
$$u(t)\leq a(t)+\int_0^t\,b(s)\cdot u(s)\,ds.$$
Notice that $a$ is nondecreasing.
If we know a priori that $\theta> 1$, then $b> 0$ and we may apply the integral form of Gronwall's inequality to conclude that
$$u(t)\leq a(t)\,\exp\left(\int_0^t\,b(s)\,ds\right).$$

We can try something similar when multiplying by $y^{-\beta}$.
We get
$$
\frac{d}{dt}\left(\frac{y^{1-\beta}}{1-\beta}  \right)\leq C_1y^{\theta-\beta}-C_2,
$$
which can be integrated to yield
\begin{align}
&\frac{y(t)^{1-\beta}}{1-\beta}-\frac{y(0)^{1-\beta}}{1-\beta}\leq -C_2t+\int_0^t\,C_1\,(1-\beta)\,y(x)^{\theta-1}\cdot\frac{y(x)^{1-\beta}}{1-\beta}\,dx\\
\iff \quad&
\frac{y(t)^{1-\beta}}{1-\beta} \leq
\left(\frac{y(0)^{1-\beta}}{1-\beta}-C_2t\right) + 
\int_0^t\,C_1\,(1-\beta)\,y(x)^{\theta-1}\cdot\frac{y(x)^{1-\beta}}{1-\beta}\,dx.
\end{align}
Then, with $u(t)=\frac{y(t)^{1-\beta}}{1-\beta}$, $a(t)=\frac{y(0)^{1-\beta}}{1-\beta}-C_2t$ and $b(t)=C_1\,(1-\beta)\,y(t)^{\theta-1}$, the inequality can be written as
$$u(t)\leq a(t)+\int_0^t\,b(s)\cdot u(s)\,ds.$$
Notice that now $a$ is nonincreasing.
If we know a priori that $\beta< 1$, then $b> 0$ and we may apply the integral form of Gronwall's inequality to conclude that
$$u(t)\leq a(t)+\int_0^t\,a(s)\,b(s)\,\exp\left(\int_a^t\,b(r)\,dr\right)\,ds.$$

This leaves the case where $\theta \leq1$ and $\beta \geq1$ without a bound.
Perhaps I will revisit this later and try and find something for these cases, or perhaps this half-answer can inspire someone to provide a more general answer.
