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
2026-03-27 10:15:45.1774606545
Proof |1/cosh(x)|≥|ln(cosh(x)/(1+cosh(x))|
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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 $y = \cosh x \ge 1$: $$ \left\lvert\ln \frac{y}{1+y}\right\rvert = - \ln \frac{y}{1+y} = \ln \frac{1+y}{y} \le \frac 1y $$