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](http://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_lemma) nor its [generalisation by Bihari](http://en.wikipedia.org/wiki/Bihari%27s_inequality), because it involves second derivatives.


I'd be glad to hear all suggestions!



Generalisations of the Gronwall's lemma
You can calculate the seconde derivative of $F^{1/p}$. you'll find that it is negative, and the seconde derivative of $G^{1/p}$ is 0, so we have $(F^{1/p})^{''}\leq (G^{1/p})^{''} $
Then by integrating twice you'll have $F^{1/p}\leq G^{1/p} $ and then $F\leq G$


