How to prove that $(1-y) \;\text{csch}\;[x (1-y)]>\text{csch}(x)$ for $0I 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.
 A: 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 $x < 0$.
A: 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.
A: Too long for comments.
At least, around integer values of $y$, series expansions shows the inequality since we have
$$\text{lhs - rhs} = (x \coth (x)-1)\, \text{csch}(x)\,y+O\left(y^2\right)$$
$$\text{lhs - rhs} =\frac{1}{x}-\text{csch}(x)+O\left((y-1)^2\right)$$
$$\text{lhs - rhs} = (x \coth (x)-1)\, \text{csch}(x)\,(2-y)+O\left((y-2)^2\right)$$
