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.

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
