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I want to prove that this inequality $$(1-y) \;\text{csch}\;[x (1-y)]>\text{csch}(x)$$ holds for $x>0$ and $0<y<2$?

Any hints and suggestions are welcome.

katy98
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  • This is equivalent to $\sinh(x)/x$ increasing for $x > 0$. That $1-y$ is a clever disguise. See my answer. – marty cohen Aug 30 '20 at 04:31
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    It suffices to prove that $x(1-y) \operatorname{csch} x(1-y) > x \operatorname{csch} x$. Let $f(u) = u \operatorname{csch} u$. We need to prove that $f(u)$ is strictly decreasing on $[0, \infty)$. Just take derivatives. – River Li Aug 30 '20 at 06:12
  • @RiverLi Thanks for your comment, this might work. – katy98 Aug 30 '20 at 10:47

3 Answers3

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Note: In response to comments, I assume that $x > 0$ and $y < 1$. I will show later (it is late here) that this will take care of the other cases.

Want $(1-y)/\sinh(x (1-y)) \gt 1/\sinh(x) $ for $0 < y < 2$ or $(1-y)\sinh(x) \gt \sinh(x(1-y)) $.

Letting $z = 1-y$, this is $z\sinh(x) \gt \sinh(zx) $ for $-1 < z < 1$.

Dividing by $zx$, this is $\dfrac{\sinh(x)}{x} \gt \dfrac{\sinh(zx)}{zx} $.

Since $\sinh(x)/x$ is even, we only need to look at $x > 0, z > 0$.

Since $f(x)=\sinh(x)/x$ is increasing (see below) and $f(0) = 1$, and $0 < z < 1$, $zx < x$ so $\dfrac{\sinh(zx)}{zx} \lt \dfrac{\sinh(x)}{x} $.

To show $f(x)=\sinh(x)/x$ is increasing for $x > 0$:

$\sinh(x) =\sum_{n=0}^{\infty} \dfrac{x^{2n+1}}{(2n+1)!} $ so $\dfrac{\sinh(x)}{x} =\sum_{n=0}^{\infty} \dfrac{x^{2n}}{(2n+1)!} =1+\sum_{n=1}^{\infty} \dfrac{x^{2n}}{(2n+1)!} $ so that $\left(\dfrac{\sinh(x)}{x}\right)' =\sum_{n=1}^{\infty} \dfrac{2nx^{2n-1}}{(2n+1)!} \gt 0$ for $x > 0$.

(Added later)

So we want $z/\sinh(xz) \gt 1/\sinh(x) $ where $|z| < 1$.

I show above this is true for $0 < z < 1$ and $x > 0$. I will now consider the other cases. The basic identity used is $\sinh(-x) = -\sinh(x) $.

If $-1 < z < 0$ and $x > 0$ then $\dfrac{z}{\sinh(xz)} =\dfrac{-z}{-\sinh(xz)} =\dfrac{-z}{\sinh(x(-z))} $ so this is the same as above.

If $-1 < z < 0$ and $x < 0$ then $\dfrac{z}{\sinh(xz)} =-\dfrac{-z}{\sinh((-x)(-z))} $ and $\dfrac1{\sinh(x)} =-\dfrac1{\sinh(-x)}$ so $z/\sinh(xz) \gt 1/\sinh(x) $ is the same as $-\dfrac{-z}{\sinh((-x)(-z))} \gt -\dfrac1{\sinh(-x)} $ or $\dfrac{-z}{\sinh((-x)(-z))} \lt \dfrac1{\sinh(-x)} $ and in this case the inequality is reversed since the signs are reversed.

If $z > 0$ and $x < 0$ then $\dfrac{z}{\sinh(xz)} =-\dfrac{z}{\sinh((-x)z)} $ and $\dfrac1{\sinh(x)} =-\dfrac1{\sinh(-x)}$ so $z/\sinh(xz) \gt 1/\sinh(x) $ is the same as $-\dfrac{z}{\sinh((-x)z)} \gt -\dfrac1{\sinh(-x)} $ or $\dfrac{z}{\sinh((-x)z)} \lt \dfrac1{\sinh(-x)} $ and in this case the inequality is reversed since the signs are reversed.

Therefore the inequality is reversed when $x < 0$.

marty cohen
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  • The problem is so simple .... after reading the nice solution ! Cheers :-) – Claude Leibovici Aug 30 '20 at 04:30
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    @martycohen Please check: $z\sinh(x) \gt \sinh(zx)$, dividing by $zx$, but $zx$ can be negative. – River Li Aug 30 '20 at 04:50
  • Since $\sinh(x)/x$ is even, we only need to look at $x>0, z>0$. – marty cohen Aug 30 '20 at 05:02
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    @martycohen But you just say this later. You should say this before, otherwise, you can not get: $z\sinh(x) \gt \sinh(zx)$ for $-1 < z < 1$, Dividing by $zx$, that is $\dfrac{\sinh(x)}{x} \gt \dfrac{\sinh(zx)}{zx}$. – River Li Aug 30 '20 at 05:07
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    Also check your first line: $(1-y)/\sinh(x (1-y)) \gt 1/\sinh(x)$ for $0 < y < 2$, or $(1-y)\sinh(x) \gt \sinh(x(1-y))$. At this moment, you do not assume $0 < y < 1$, you can not get the latter. – River Li Aug 30 '20 at 05:12
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If $y \ne 1,$ then $y = 1 \pm t,$ where $t > 0,$ and the inequality to be proved is equivalent to $$ \frac{\sinh(tx)}{tx} < \frac{\sinh x}x \quad (x > 0, \ 0 < t < 1). $$ An equivalent statement is that the function $$ g(x) = \frac{\sinh x}x \quad (x \ne 0) $$ is strictly increasing for $x > 0.$

We have $$ \lim_{x \to 0}g(x) = \lim_{x \to 0} \frac12\left(\frac{e^x - 1}x + \frac{1 - e^{-x}}x\right) = 1, $$ so, if $g(x)$ is strictly increasing for $x > 0,$ it follows that: $$ g(x) > 1 \quad (x > 0), $$ and therefore: $$ \lim_{y \to 1}((1 - y)\operatorname{csch}(x(1 - y))) = \lim_{t \to 0}\frac{t}{\sinh(tx)} = \lim_{t \to 0}\frac1{xg(tx)} = \frac1x > \operatorname{csch} x \quad (x > 0), $$ which gives a reasonable interpretation to the stated inequality in the strictly meaningless case $y = 1.$

As for the main case: $$ g'(x) = \frac{x\cosh x - \sinh x}{x^2} = \frac{(\cosh x)(x - \tanh x)}{x^2} > 0 \quad (x > 0). $$ This follows from the first of the two inequalities (2) that were proved in my answer to the OP's more recent question Any idea to prove that $\coth (x)-(1-y) \coth [x (1-y)]<\frac{4 x}{1-x^4}$?

Repeating - and slightly extending - that proof here, to make this answer self-contained:

The graph of the convex function $u \mapsto 1/u$ lies above the tangent at $(1, 1),$ therefore $$ 2x < \int_{1 - x}^{1 + x}\frac{du}u = \log\frac{1 + x}{1 - x} \quad (0 < x < 1), $$ therefore $$ (1 - x)e^{2x} < 1 + x \quad (x > 0), $$ because this inequality holds trivially if $x \geqslant 1.$

Therefore, for all $x > 0,$ $e^{2x} - 1 < x(e^{2x} + 1).$ Equivalently, $$ \tanh x < x \quad (x > 0), $$ which gives $g'(x) > 0,$ as required.

Calum Gilhooley
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Too long for comments.

At least, around integer values of $y$, series expansions shows the inequality since we have $$\text{lhs - rhs} = (x \coth (x)-1)\, \text{csch}(x)\,y+O\left(y^2\right)$$ $$\text{lhs - rhs} =\frac{1}{x}-\text{csch}(x)+O\left((y-1)^2\right)$$ $$\text{lhs - rhs} = (x \coth (x)-1)\, \text{csch}(x)\,(2-y)+O\left((y-2)^2\right)$$

Claude Leibovici
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