Nonlinear Gronwall inequality Suppose a (continuous, non-negative) function $f$ satisfies
$$ f'(t) \leq f(t)^2 $$
for $t \in [0,1]$. Set 
$$ g(t) = \exp\left(-\frac{1}{f(t)}\right). $$
Then
$$ g'(t) = \frac{f'(t)}{f(t)^2} g(t) \leq g(t), $$
so applying the standard Gronwall inequality one has $g(t) \leq g(0) e^t$. Since $\log$ is an increasing function, applying $\log$ to both sides and rearranging gives
$$ f(t) \leq \left( \frac{1}{1-tf(0)} \right) f(0). $$
Thus if $f(0) \leq \epsilon$ for $\epsilon >0$ small enough, we get $f(t) \leq C f(0)$ for all $t \in [0, 1]$. This proves a kind of small data nonlinear Gronwall inequality, but only starting from the differential form of the inequality.

My question: Suppose instead that we have 
  $$ f(t) - f(0) \leq \int_0^t f(s)^2 \, ds. $$
  Can we deduce the same result?

 A: Here is an easy way to get the first part: $$\color{red}{f'(t) \leq f(t)^2} \Longleftrightarrow \frac{f'(t) }{ f(t)^2}\le 1 \Longleftrightarrow  \left(-\frac{1}{f(t)}\right)'\le 1$$
integrating this latest gives 
$$\left( -\frac{1}{f(t)}+\frac{1}{f(0)}\right)\le t\Longleftrightarrow  f(t) \leq \left( \frac{1}{1-tf(0)} \right) f(0).$$

Now if you have, $$f(t) - f(0) \leq \int_0^t f(s)^2 \, ds\implies \color{blue}{f^2(t)\le \left(f(0) + \int_0^t f(s)^2 \, ds\right)^2} \\\implies g'(t) \le g^2(t)$$
  where we set $$g(t) = f(0)+\int_0^t f(s)^2 \, ds\implies   g'(t)= f^2(t)$$
  This is  of the same type as the previous inequality.

A: You get faster to the final inequality of the first part by just dividing by $f(t)^2$, as then
$$
\frac1{f(0)}-\frac1{f(t)}=\int_0^t\frac{f'(s)}{f(s)^2}\,ds\le\int_0^t 1\,ds=t
$$
so that indeed
$$
f(t)\le \frac{f(0)}{1-tf(0)}
$$

For the second part, the question itself, consider the function $$g(t)=-\frac{1}{f(0)+\int_0^tf(s)^2 ds}.$$ 
Then 
$$
g'(t)=\frac{f(t)^2}{\left(f(0)+\int_0^tf(s)^2 ds\right)^2}\le 1
$$
so that
$$
g(t)\le -\frac1{f(0)}+t
$$
and finally again
$$
f(t)\le -\frac1{g(t)}\le \frac{f(0)}{1-tf(0)}
$$
