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
 A: It is false if there is no constraints on the limit of the derivative when $t$ goes to $0$. Let $f(t) = \sqrt t$. We see that $f(t)$ is an increasing function. Then $f'(t) = \frac 1{2\sqrt t}$ and since $f(t)$ is bounded in the interval say $[0,1]$, for any constant $C$, $Ce^{f(t)}$ is bounded in $[0,1]$. But $f'(t)$ is not, so this is not possible.
Maybe the actual question puts some more constraints on $f$ : here I used the fact that it is possible that the derivative of $f$ goes to infinity as $t \to 0$. Perhaps you might ask for the derivative to have a limit when $t \to 0^+$?
Hope that helps,
A: It is false as can be seen by the following proof by contradiction.  Suppose it is true, and such a $C$ and $T$ exist.  Then consider the following sequence of functions $f_n(t)=n(t-T) + 1+2T$ for $t\geq T$ and then extend $f_n$ smoothly for $t < T$ while keeping $f_n > 1$ and  $f_n'\geq 1$.  Then we have that $f_n'(T) \leq Ce^{f(T)}$, and $f_n'(T)=nT$ and $e^{f(T)}=e^{1+2T}$, hence the inequality would say that $nT \leq Ce^{1+2T}$ or $C \geq nTe^{-(1+2T)}$ which is a contradiction as $n\to\infty$.
NOTE:  I assumed that you do not want the constants to depend on $f$ (as this would change the problem).
A: 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.
