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.

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.
