By looking at the graph of RHS-LHS, I believe the following inequality holds:
$$\coth x\;\leq\; x^{-1}+x \quad\text{for } x>0$$
I can't think of a way to prove it right now, and I would love a hint and/or reassurance that it's true.
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1Perhaps compute the series expansion of $\coth x - (1/x + x)$ and see if the terms show that it is always negative. – David G. Stork Mar 26 '19 at 19:54
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That's the first thing I thought of but seems messy. Will do if there's no other option. Thank you for your comment! – folouer of kaklas Mar 26 '19 at 19:55
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Note that $x+x^{-1}-\coth x$ has derivative $\coth^2 x-x^{-2}\ge 0$, while $\lim_{x\to 0^+}x^{-1}-\coth x=\lim_{x\to 0^+}(-x/3)=0$.
J.G.
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From the series expansion of the exponential function, if $x > 0$, then: $$ \frac{e^{2x}-1}{2} > x + x^2, $$ therefore: $$ \coth x = 1 + \frac{2}{e^{2x}-1} < 1 + \frac{1}{x + x^2} = 1 + \frac{1}{x} - \frac{1}{1 + x} = \frac{x}{1 + x} + \frac{1}{x} < x + \frac{1}{x}. $$
Calum Gilhooley
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