Let $f:[0,\alpha]\to \mathbb{R}$ solution of a Cauchy problem , prove $\alpha <3.$ Let $f:[0,\alpha] \to \mathbb{R}$ be a solution of the Cauchy problem 
$$ 
\begin{cases}
f'(t)=f(t)^2+t, \\
f(0)=0.
\end{cases}
$$
Prove that $\alpha <3.$
I know that I should try to prove that $f$ blows up to $+\infty$ before time $t=3$. I tried to apply the comparison theorem, but I cound't find a suitable function to use. Any help is appreciated. 
 A: First, note that $f'(t) \ge t$, so $f(t) \ge \frac 12 t^2$, in particular $f(1) \ge \frac 12$. 
Second, $f$ can be compared with the solution $g$ of 
$$ \begin{cases} g'(t) = g(t)^2, \\ g(1) = \frac{1}{2}. \end{cases} $$ 
Since $f'(t) \ge f(t)^2$ and $f(1) \ge g(1)$, we have $f(t) \ge g(t)$ for $t \ge 1$. 
But $g$ can be calculated to be $g(t) = \frac{1}{3-t}$, so $f$ blows up at time $t=3$ or earlier. 
A: For any $t>0$ we have that $f'(t)$ is positive, hence $f(t)$ is increasing and invertible. Let $g=f^{-1}$.
We have $f(g(t))=t$, hence 
$$ t^2+g(t) = f'(g(t)) = \frac{1}{g'(t)} $$
and from the positivity of $g(t)$ we may deduce $g'(t)\leq\frac{1}{g(t)}$, from which $g(t)\leq\sqrt{2t}$.
Let $t_0$ be the least time such that $g(t_0)\geq 1$. By the previous inequality, $t_0\geq\frac{1}{2}$.
From that point on, $g'(t)\leq\frac{1}{t^2+1}$, hence $g(t)$ is bounded by $1-\arctan t_0+\frac{\pi}{2}$.
In particular the life span of $f$ cannot exceed $1-\arctan\frac{1}{2}+\frac{\pi}{2}<\color{red}{2+\frac{1}{9}}$.
