Inequality for a gamma function Let $s=\sigma+it$ and $\Gamma(s)$ be the Euler gamma function. 
Does the inequality hold?
$$
\left|\frac{\Gamma(s)}{\Gamma(2-s)}\right|\leq |s|^{2(\sigma-1)},\, 1<\sigma<2,\,  t\in \mathbb{R}.
$$
Difficulties to prove inequality appears when $\sigma$ approximates 1.
Such inequality appeared studying a zeta functions of a second order. Namely, comparing the values of the Selberg zeta-function for the modular subgroup $PSL(2,\mathbb{Z})$ across the critical line: |Z(1-s)|>|Z(s)| (|Z(1-s)|<|Z(s)| ?), $1/2<\sigma<1$.
We can show that
$$
\left|\frac{\Gamma(s)}{\Gamma(2-s)}\right|\leq \left|s-2-\frac{\sqrt{2}}{2}\right|^{2(\sigma-1)},\, 1<\sigma<2,\, t\in \mathbb{R}.
$$
See the Lemma 7 in http://link.springer.com/article/10.1007%2Fs00025-015-0486-7#page-1
However, the same technic doesn't work for the first inequality.
 A: *

*I am not sure this estimate is true. In the cited preprint signs of s are the same, yours are opposite.

*Standard inequalities gives not power but exponential growth
$$
|\frac{\Gamma(s)}{\Gamma(2-s)}|\le \frac{1}{\pi} \sinh(\pi |s|).
$$ 
Really the better estimate is true? 

A: There is a genuine asymptotic $\Gamma(s+a)/\Gamma(s)\sim s^a$ for $s$ in a half-plane to the right of $0$, as $|s|\to \infty$, for bounded $a$. (This is proven in many places, as a corollary of Watson's lemma, much easier than Stirling-Binet-Laplace, and not using the latter. E.g., in course notes at http://www.math.umn.edu/~garrett/m/mfms/ named "asymptotics of integrals, including the Gamma function".
Then use $|\Gamma(\sigma-it)|=|\Gamma(\sigma+it)|$, so $|\Gamma(2-s)|=|\Gamma(2-\sigma+it)|$, and $|\Gamma(\sigma+it)/\Gamma(2-\sigma+it)|\sim |t|^{2\sigma-2}$. An asymptotic, not an inequality.
A: Write $s=\sigma+it$.  Let $\ell(\sigma)$ be the line such that $\ell(1)=1$ and $\ell(2)=0$, namely $\ell(\sigma)=2-\sigma$.  On the line $\sigma = 1$, we have
$|\frac{\Gamma(s)}{\Gamma(2-s)}| = |\frac{\Gamma(1+it)}{\Gamma(1-it)}|=1=:M_1$.
On the line $\sigma = 2$, we have (by $s\Gamma(s)=\Gamma(s+1)$)
$|\frac{\Gamma(s)}{\Gamma(2-s)}| = |\frac{\Gamma(2+it)}{\Gamma(-it)}| = |t|\cdot|t+i|=:M_2$.
The Phragmen-Lindelof principle now tells us that for $1\leq \sigma\leq 2$, we have
$|\frac{\Gamma(s)}{\Gamma(2-s)}|\leq M_1^{\ell(\sigma)} M_2^{1-\ell(\sigma)}=(|t|\cdot|t+i|)^{\sigma-1}$.
This is (almost) your desired inequality.
