From page 592 of Boyd & Vandenberghe's Convex Optimization,
Using concavity of the logarithm, we also have $$ \log(1 + 1/\sqrt{m}) \ge (\log 2)/\sqrt{m} \tag{1}$$
How can we make use of the concavity of the logarithm to get $(1)$? Actually, we can replace $1/\sqrt{m}$ with $x$ to simplify our notations and get
$$
\log(1 + x) \ge (\log 2)x
$$
This looks easy. I wasted a few hours on this problem, but failed to prove it.
In the original post I omitted the context in the book. Sorry about that. Specifically, $m$ denotes the number of inequalities which implies $0<1/\sqrt{m}\le 1$ and $0<x\le 1$. Now everything is clear.
Best Answer
For concave function, we have $$f(\lambda x + (1-\lambda)y) \ge \lambda f(x) + (1-\lambda ) f(y)$$
Hence if $m \ge 1$,
\begin{align} \log \left( 1+\frac1{\sqrt{m}}\right) &= \log \left( 1-\frac1{\sqrt{m}}+\frac{2}{\sqrt{m}}\right)\\ &=\log \left( \left( 1-\frac1{\sqrt{m}}\right)1+\left(\frac{1}{\sqrt{m}}\right) 2\right)\\ &\ge \left( 1-\frac1{\sqrt{m}}\right)\log 1 + \left(\frac{1}{\sqrt{m}}\right) \log 2\\ &= \left(\frac{1}{\sqrt{m}}\right) \log 2 \end{align}