Using an inequality on derivative of a function to arrive at an inequality on the function

Question

Given a function $f \in C^1((0,T])$ for some $T>0$ , satisfying inequality \begin{equation} |f'(t)| \leq \frac{C}{t} \text{ for some }C>0 \end{equation} I need to prove that \begin{equation} |f(t)| \leq C'\log\Big(1+\frac{1}{t}\Big) \text{ for some }C'>0. \end{equation} My try is as follows \begin{align} |f(t)|-|f(T)| &\leq |f(T)-f(t)| \\ & = \Big| \int_{t}^{T}f'(t)dt \Big| \\ & \leq \int_{t}^{T} \frac{C}{t}dt \\ & = C\log\Big(\frac{T}{t}\Big) \end{align} This gives \begin{equation} |f(t)| \leq C''+C\log\Big(\frac{1}{t}\Big) \text{ where }C''=|f(T)|+C\log(T). \end{equation} Taking $C'=max\{C'',C\}$ \begin{equation} |f(t)| \leq C'\Big\{1+\log\Big(\frac{1}{t} \Big)\Big\} \end{equation} which not same as what I need to show.

Thanks in advance.

Answer 1

$$1 + \log \frac{1}{t} \le 1 + \log(1 + \frac{1}{t}) \le c \log(1 + \frac{1}{t})$$ for $c = \frac{1}{\log(1 + \frac{1}{T})} + 1$.