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

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).
