Show that $\frac{(1-y^{-1})^2}{(1-y^{-1 -c})(1-y^{-1+c} )}$ is decreasing in $y > 1 $. I am interested in the function
$f(y) = \frac{(1-y^{-1})^2}{(1-y^{-1 -c})(1-y^{-1+c} )},$
for values of $c \in (0,1)$, and $y > 1$, and have been trying to show that the function is decreasing.
I have tried differentiating the function, but this does not yield a particularly amenable formula to decide whether $f'(y) < 0$. Similarly I have considered the ratios $f(y)/f(\tilde y)$, but again have been unable to ascertain an inequality for this.
I'd appreciate any other approaches to the problem.
 A: EDIT
An easier proof:
Note
\begin{eqnarray}
f(y) &=& \frac{(1-y^{-1})^2}{(1-y^{-1-c})(1-y^{-1+c})} \\
     &=& \frac{(y^{1/2}-y^{-1/2})^2}{(y^{\frac{1+c}{2}}-y^{-\frac{1+c}{2}})(y^{\frac{1-c}{2}}-y^{-\frac{1-c}{2}})} \\
&=& \left(\frac{\sinh x}{\sinh \ (1+c)x}\right)^2,
\end{eqnarray}
where $x = \log\sqrt{y}$. Thus, it's sufficient to show
$$
t(x) =\frac{\sinh x}{\sinh \ (1+c)x}
$$
is decreasing for $x>0$. Note
\begin{eqnarray}
t'(x) &=& \frac{\sinh (1+c)x \cosh x - (1+c)\sinh x \cosh(1+c)x}{\sinh^2 (1+c)x} \\
&=& \cosh x \cosh(1+c)x \frac{\tanh (1+c)x - (1+c)\tanh x}{\sinh^2 (1+c)x}.
\end{eqnarray}
All factors here are strictly positive for $x>0$ except $s(x)\equiv\tanh (1+c)x - (1+c)\tanh x$, which is strictly negative. So see this, note
\begin{eqnarray}
s'(x) = (1+c) \cosh^{-2} (1+c) x - (1+c) \cosh^{-2} x
\end{eqnarray}
and use the fact that $\cosh x$ is strictly increasing for $x>0$.

ORIGINAL PROOF (Adds nothing over one above)
You are interested in the function
$$
f(y) = \frac{(1-y^{-1})^2}{(1-y^{-1-c})(1-y^{-1+c})}
$$
for $y>1$. Note that $f(y)$ is strictly decreasing for $y>1$ iff $x\mapsto \frac{(1-x)^2}{(1-x^{1+c})(1-x^{1-c})}$ is strictly increasing for $0<x<1$ (because $s\mapsto \frac{1}{s}$ is strictly decreasing on $s>0$). Observe
$$
\frac{(1-x)^2}{(1-x^{1+c})(1-x^{1-c})} = \frac{1}{1 - \frac{x^{1+c} + x^{1-c} - 2 x}{(1-x)^2}}.
$$
The right hand side can be written $1/(1-g(x))$ where
$$
g(x) = \frac{x^{1+c} + x^{1-c} - 2 x}{(1-x)^2}.
$$
We want to show that $1/(1-g(x))$ is strictly increasing for $0<x<1$. This will follow if we are able to show $g(x)$ is positive and  strictly increasing for $0<x<1$. One can write
$$
g(x) = \frac{x}{(1-x)^2} (x^c + x^{-c} - 2).
$$
The two factors can be rewritten as follows:
\begin{eqnarray}
\frac{x}{(1-x)^2} &=& \frac{1}{(x^{1/2} - x^{-1/2})^2} \\
x^c + x^{-c} - 2 &=& (x^{c/2} - x^{-c/2})^2.
\end{eqnarray}
We end up with the lovely expression for $g(x)$:
$$
g(x) = \left(\frac{x^{c/2} - x^{-c/2}}{x^{1/2} - x^{-1/2}}\right)^2.
$$
It is now obvious $g(x) > 0$.
Now to show $g$ is strictly increasing.
We can again use the "increasing functions of increasing functions are increasing" trick to ignore the square. We also substitute $s=-\frac{1}{2}\log x$. With this transformation, $0<s<+\infty$, and since $s$ is strictly decreasing as a function of $x$, we must now show $h(s)$ is strictly decreasing for $0<s$ where
$$
h(s) = \frac{\sinh(cs)}{\sinh s}.
$$
($h$ is obtained simply by substituting $s=-\frac{1}{2}\log x$ in the $g(x)$ expression above.) 
Compute:
\begin{eqnarray}
h'(s) &=& \frac{c \sinh s \cosh (cs) - \sinh(cs) \cosh s}{\sinh^2 s} \\
&=& \cosh s\cosh(cs)\frac{c\tanh s - \tanh(cx) }{\sinh^2 s}.
\end{eqnarray}
The $\cosh$ and $\sinh$ factors are positive for $s>0$. To see that
$$
p(s) = c\tanh s - \tanh(cx)
$$
is negative, note that $p(0)=0$ and $p'(s) = c(\cosh^{-2} s - \cosh^{-2} (cs))$. Since $\cosh s$ is strictly increasing for $s>0$, $p'(s)<0$, so $h(s)$ is indeed strictly decreasing in $s$. We are done.
A: Let $f_c(x)=\frac{(1-x^{-1})^2}{(1-x^{-1-c})(1-x^{-1+c})}$. Then $\frac{\partial}{\partial c}(\ln f_c(x))=-\ln x(\frac{1}{x^{c-1}-1}+\frac{1}{x^{c+1}-1}+1)$, so since $f_0(x)=1$, it suffices to show that $\frac{1}{x^{c-1}-1}+\frac{1}{x^{c+1}-1}$ is increasing with respect to $x > 1$. $\frac{1}{x^{c-1}-1}$ is negative and increasing, $\frac{1}{x^{c+1}-1}$ is positive and decreasing, and $\frac{\partial (x^{c-1}-1)^{-1}}{\partial x} = \frac{c-1}{c+1}\left(\frac{x^{c+1}-1}{x-x^c}\right)^2 \frac{\partial (x^{c+1}-1)^{-1}}{\partial x}$, so we must show that $\frac{x^{c+1}-1}{x-x^c} > \sqrt{\frac{1+c}{1-c}}$. By L'Hopital's rule $\frac{x^{c+1}-1}{x-x^c}$ approaches $\frac{1+c}{1-c} > \sqrt{\frac{1+c}{1-c}}$ at $x=1$, so it suffices to show that $\frac{x^{c+1}-1}{x-x^c}$ is increasing on $x>1$.
$\frac{\partial}{\partial x}(\ln\frac{x^{c+1}-1}{x-x^c}) = \frac{cx^c(x^2-1)-x(x^{2c}-1)}{x(x-x^c)(x^{c+1}-1)}$, so since the denominator is positive, we must show that $cx^c(x^2-1)>x(x^{2c}-1)$. Since both sides are zero when $c=0$, taking $\frac{\partial}{\partial c}$ of both sides and simplifying shows that it is sufficient to show that $\phi_c(x) > \psi_c(x)$, where $\phi_c(x)=x^{1-c}-x^{-1-c}$ and $\psi_c(x)=\frac{2}{c + (\ln x)^{-1}}$. Now $\frac{\partial \phi_k}{\partial c}(x)=-\phi_k(x)\ln x$ and $\frac{\partial \psi_k}{\partial c}(x) = -\frac{1}{2}(\psi_k(x))^2$, so since $\ln x > \frac{1}{2}\psi_k(x)$, if $\phi_c(x) > \psi_c(x)$ for a given $c$ then $\frac{\partial \phi_c}{\partial c}(x) < \frac{\partial \psi_c}{\partial c}(x)$. Therefore it suffices to show that $\phi_1(x) > \psi_1(x)$, i.e. that $x^2-1+\frac{x^2-1}{\ln x} > 2$.
By L'Hopital's rule $x^2-1+\frac{x^2-1}{\ln x}$ approaches $2$ at $x=1$, so it suffices to show that $\frac{x^2-1}{\ln x}$ is increasing. Now $(\frac{x^2-1}{\ln x})'=\frac{x^2(2\ln x - 1)+1}{x\ln^2 x}$. Since the derivative of $x^2(2\ln x - 1)+1$ is $4x\ln x$, whose only zero (for $x > 0$) is at $x=1$,  it follows that the minimum value of $x^2(2\ln x - 1)+1$ is at $x=1$ where it equals $0$, and thus for $x>1$ we have $\frac{x^2(2\ln x - 1)+1}{x\ln^2 x} > 0$. Therefore $\frac{x^2-1}{\ln x}$ is increasing for $x>1$. QED
A: We want to show that $$ f(y) = \dfrac {\left( 1 - y^{-1} \right)^2}{\left( 1 - y^{-1-c} \right) \cdot \left( 1 - y^{1+c} \right)} $$ is decreasing in $ y > 1 $. Indeed, we can differentiate. It suffices to show that $ f'(y) < 0 $, for $ y > 1 $, if $ c \in (0, 1) $. 
So, we differentiate using the Quotient Rule: $$ f'(y) = \dfrac {(1-y^{-1-c})(1-y^{1+c})(2)(1-y^{-1})(y^{-2}) - (1-y^{-1})^2\left((1-y^{-1-c})(1-y^{1+c})\right)'}{\left( 1 - y^{-1-c} \right)^2 \cdot \left( 1 - y^{1+c} \right)^2}. $$Now, let $h(x)$ be the numerator of this function. That is: $$ h(y) = (1-y^{-1-c})(1-y^{1+c})(2)(1-y^{-1})(y^{-2}) - (1-y^{-1})^2\left((1-y^{-1-c})(1-y^{1+c})\right)'. $$Since the denominator of $f'(y)$ is always positive, $ f'(y) < 0 \implies h(y) < 0 $. 
Let $ g(y) = \left( 1 - y^{-1-c} \right) \cdot \left( 1 - y^{1+c} \right) $. 
Then, $$ g(y) = 1 - y^{-1-c} - y^{1+c} + 1 = 2 - y^{1+c} - y^{-1-c}, $$ so $$ g'(y) = - \left( 1+c \right) y^c - \left( -1 - c \right) y^{-2-c} = - \left( 1 + c \right)y^c + (1+c) y^{-2} - c, $$ $$ \implies g'(y) = (1+c) \cdot \left( y^{-2-c} - y^c \right). $$Now, going back to $h(y)$, we have: $$ (1-y^{-1-c})(1-y^{1+c})(2)(1-y^{-1})(y^{-2}) - (1-y^{-1})^2(1+c)(y^{-2-c}-y^c) < 0. $$
$ \implies (1-y^{-1-c})(1-y^{1+c})(2)(1-y^{-1})(y^{-2}) < (1-y^{-1})^2 (1+c)(y^{-2-c} - y^c) $ 
$ \stackrel{y \ne 1}{\implies} (2 - y^{-1-c} - y^{1+c})(2)(y^{-2}) < (1-y^{-1})(1+c)(y^{-2-c}-y^c) $
$ \implies 2 \cdot \left( 2y^{-2} - y^{-3-c} - y^{-1+c} \right) < (1+c) \cdot (y^{-2-c} - y^c - y^{-3-c} + y^{-1+c}) $
$ \implies 4y^{-2} - 2y^{-3-c} - 2y^{-1+c} < y^{-2-c} - y^c - y^{-3-c} + y^{-1+c} + cy^{-2-c} - cy^c - cy^{-3-c} + y^{-1+c}) $
From here, group like terms and we get that, if $ c \in (0, 1) $, the result is true for $ y > 1 $. 
