Is this inequality true? $\coth x\leq x^{-1}+x$

Note that $x+x^{-1}-\coth x$ has derivative $\coth^2 x-x^{-2}\ge 0$, while $\lim_{x\to 0^+}x^{-1}-\coth x=\lim_{x\to 0^+}(-x/3)=0$.


From the series expansion of the exponential function, if $x > 0$, then: $$ \frac{e^{2x}-1}{2} > x + x^2, $$ therefore: $$ \coth x = 1 + \frac{2}{e^{2x}-1} < 1 + \frac{1}{x + x^2} = 1 + \frac{1}{x} - \frac{1}{1 + x} = \frac{x}{1 + x} + \frac{1}{x} < x + \frac{1}{x}. $$

Tags:

Inequality

Calculus

Real Analysis

Hyperbolic Functions

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