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
Proof |1/cosh(x)|≥|ln(cosh(x)/(1+cosh(x))|
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real-analysis
inequality
logarithms
absolute-value
hyperbolic-functions
1 Answers
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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 $y = \cosh x \ge 1$: $$ \left\lvert\ln \frac{y}{1+y}\right\rvert = - \ln \frac{y}{1+y} = \ln \frac{1+y}{y} \le \frac 1y $$
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0I don't see how your conclusion follows from ln(1+x)≤x for x>-1. can you elaborate? – 2017-02-20
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1@McRaylie: $\ln \frac{1+y}{y} = \ln \left( 1 + \frac 1y \right) \le \frac 1y$. – 2017-02-20