Show that $\frac{1}{\cosh(x)} + \log\left(\frac{\cosh(x)}{1+\cosh(x)}\right) \ge 0$ I want to show that the following is true: > $$\frac{1}{\cosh(x)} + \log(\frac{\cosh(x)}{1+\cosh(x)}) \ge 0$$ Below is how I approached it, but I did not end up with an acceptable answer. \begin{align} \frac{1}{\cosh(x)} + \log\left(\frac{\cosh(x)}{1+\cosh(x)}\right) & = \log\left(e^{\frac{1}{\cosh(x)}}\right) + \log\left(\frac{\cosh(x)}{1+\cosh(x)}\right) \\\ & = \log\left(\frac{e^{\frac{1}{\cosh(x)}} \cdot \cosh(x)}{1+\cosh(x)}\right). \end{align} I don't know where to go from here so that I can show that this non-negative.

A slicker approach is to rewrite the identity as $$s:=\operatorname{sech}x\ge-\ln\frac{\cosh x}{1+\cosh x}=\ln\frac{1+\cosh x}{\cosh x}=\ln (1+s),$$which is trivial from the famous inequality $\exp y\ge 1+y$ with $y:=\ln (1+s)$.