I am reading a book about Euler equation and at some point we aim to prove Beale-Kato-Majda theorem. In the proof, the author uses that $$\log_+ \frac{a}{b} \le \log(1 + a) + 1,$$ for $a, b > 0$ and $\log_+ x$ equal to $0$ up to $1$ and $\log(x)$ afterwards. This is surely a typo as for $b \to 0$ the left hand side blows up while the right hand side remains bounded. For the rest of the computation in the proof, it seems that the right hand side should be $\log(1 + a) + 1/b$ instead. However, I struggle to prove it.
For $a \le b$, the left hand side vanishes so the inequality is obvious. Now for $a > b$, it amounts to proving that $$\log\left(\frac{a}{b(1 + a)}\right) \le \frac{1}{b}.$$ I've tried many things but I admit that I couldn't find anything convincing. Any idea?
Since $f(t)=\log(t)$ is concave, we have the inequality $$ \log(a/b) = f(a/b)\leq f(1+a)+f'(1+a)(a/b - 1 - a)= \log(1+a)+\frac{a/b-1-a}{1+a}=\log(1+a)+\frac{1}{b}\frac{a}{(1+a)} - 1\leq\log(1+a)+\frac{1}{b}, $$ for $a/b \geq 1$.