Suppose we have the following differential inequality
$F''(w)\le \frac{p-1}{p}\frac{(F'(w))^2}{F(w)}$ on $w\in(w_0,w_0+\varepsilon)$, $w_0>0$, $\varepsilon>0$, $p>1$. In addition, $F(w)>0$, $F'(w)>0$ for $w\in[w_0,w_0+\varepsilon)$.
Let's consider the solution of the Cauchy problem
$\begin{cases}G''(w)= \frac{p-1}{p}\frac{(G'(w))^2}{G(w)},& w\in(w_0,w_0+\varepsilon),\\G'(w_0)=F'(w_0),&\\G(w_0)=F(w_0),&\end{cases},$ which can be written explicitly:
$G(w) = F(w_0)\left(1+\frac{F'(w_0)}{F(w_0)}\frac{w-w_0}{p}\right)^p$.
Is it possible to compare the functions $F$ and $G$ in any meaningful way? This case fits neither the canonical version of the Gronwall's lemma nor its generalisation by Bihari, because it involves second derivatives.
I'd be glad to hear all suggestions!
For the sake of having an answer, I repost here a solution based on the idea proposed by Hicham at mathoverflow.
In fact, the first differential inequality leads to $\frac{d^2}{dw^2}((F(w))^{1/p})\le 0$, or $(F(w))^{1/p}$ being concave. In the same time, $(G(w))^{1/p}$ is a tangent line to $(F(w))^{1/p}$ in $w_0$, therefore $(F(w))^{1/p}\le (G(w))^{1/p}$ and $ F(w) \le G(w) $.