<s>[INST] 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.
 [/INST] 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 $. 
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