real analysis - Proof |1/cosh(x)|≥|ln(cosh(x)/(1+cosh(x))|

I am doing a problem where I need to prove that $$\left|\frac{1}{\cosh(x)}\right|≥\left|\ln\left(\frac{\cosh(x)}{1+\cosh(x)}\right)\right|$$
Without differentiation, and I can't find a way to prove it. Can anyone prove it without

Answer

Using
$$
\ln (1+x) \le x \text { for } x > -1
$$ (see for example how to prove that $\ln(1+x)< x$) the following holds for $y = \cosh x \ge 1$:
$$
\left\lvert\ln \frac{y}{1+y}\right\rvert = - \ln \frac{y}{1+y} = \ln \frac{1+y}{y} \le \frac 1y
$$