Let $f\colon [0,\infty)\to [0,\infty)$ be a smooth function such that $f(0)=0$ and \begin{equation} f'(t) \le a(t)+b(t) \bigl( f(t) \bigr)^\alpha \label{eq:1} \tag{1} \end{equation} for all $t\ge 0$, some $\alpha\in(0,1)$ and some functions $a,b\colon[0,\infty)\to(0,\infty)$.
Question: What (possibly sharp) bound on $f$ can be derived in terms of $a,b$ and $\alpha$? For my problem, we can wlog. assume that $a$ and $b$ are non-decreasing, but the answer itself might be interesting in generality.
My attempt
If $a\equiv 0$, then we can reduce the problem to the linear setting and apply the standard Gronwall lemma. Namely, \eqref{eq:1} is then equivalent to $(f^{1-\alpha}(t))'\le (1-\alpha)b(t)$, whence \begin{equation}f(t)\le \bigl( (1-\alpha)\int_0^t b(s)\,ds \bigr)^{1/(1-\alpha)}.\label{eq:2}\tag{2}\end{equation}
Also, we could estimate brutally $x^\alpha\le x$ for $x\ge 1$ and $f(t)\le f(t)+1$ to get that \eqref{eq:1} implies $f'(t) \le a(t)+b(t)+b(t)f(t)$, whence by the Gronwall lemma $$ f(t) \le \int_0^t (a(s)+b(s))e^{\int_s^t b(u)\,du}\,ds. $$ This estimate however does not take $\alpha$ into considerations and seems way off (compare e.g. with \eqref{eq:2} in case $a\equiv 0$).