How to prove such a simple inequality? Let $f\in C[0,\infty)\cap C^1(0,\infty)$ be an increasing convex function with $f(0)=0$ $\lim_{t->\infty}\frac{f(t)}{t}=+\infty$ and $\frac{df}{dt} \ge 1$. 
Then there exists constants $C$ and $T$ such that for any $t\in [T,\infty)$, $\frac{df}{dt}\le Ce^{f(t)}.$
Is it correct? If the conditions are not enough, please add some condition and prove it. Thank you

Here are some more details on my answer in the comment above.  Let
$$ g_n(t) = 2^{-n}\left(\frac{1}{\pi}\arctan\left(2^ne^{4n}\pi(t-n)\right) + \frac{1}{2}\right).$$
Note that $|g_n(t)| \leq 2^{-n}$ and $g_n'(t) \geq 0$ for all $t$ and in particular $g_n'(n) = e^{4n}$.
Now define
$$f(t) = t + 2 + \sum_{n=1}^\infty g_n(t).$$
You should verify that this series actually converges and $f\in C^1$ (this is not hard).  Then $|f(t)| \leq t + 3$ and $f'(n) = 1 + e^{4n}$.  If the statement were true, then there would exist $C$ such that
$$e^{4n} \leq Ce^{n+3},$$
for all integers $n$ which is a contradiction.
