Let us consider the function $$ x \mapsto \left| \frac{\Gamma(x+z+iy)}{\Gamma(x+iy)} \right|^2, $$ where $x,y,z \in \mathbb R$, $x \ge 1/2$, $z >0$, $i$ is the imaginary unit, and $\Gamma$ is the Euler gamma function.
Is this function non-decreasing?
My straightforward attempt to check this would be by computing the derivative, but this leads me to some impenetrable formulas.
Numerical evidence also suggests that the answer to the question is yes.
On
By Gary's comment, it is enough to show that for any $y,C,z\in\mathbb{R}^+$ the function
$$ f:x \mapsto \frac{(x+C)^2+y^2}{(x+C+z)^2+y^2} $$
is increasing on $\left[\frac{1}{2},+\infty\right)$. We may notice that
$$ \frac{f'(x)}{f(x)}=\frac{d}{dx}\left(\log f(x)\right) = \frac{2(C+x)}{(C+x)^2+y^2}-\frac{2(C+x+z)}{(C+x+z)^2+y^2}=2z\frac{(C+x)(C+x+z)-y^2}{((C+x)^2+y^2)((C+x+z)^2+y^2)}$$ so the monotonicity clearly holds if $x\geq |y|$, but it might fail otherwise.
On
Some thoughts:
By Gary's comment, letting $$f(x) := \ln \left| {\frac{{\Gamma (x + z + iy)}}{{\Gamma (x + iy)}}} \right|^2,$$ we have $$f'(x) = \sum_{k=0}^\infty \left(\frac{2(k+x)}{(k+x)^2+y^2} - \frac{2(k+x+z)}{(k+x+z)^2+y^2}\right).$$
If $x^2 + xz \ge y^2$, we have $\frac{2(k+x)}{(k+x)^2+y^2} - \frac{2(k+x+z)}{(k+x+z)^2+y^2} \ge 0$ and $f'(x) \ge 0$.
By the answers of Gary and River Li, it suffices to show that $$ U(x,z)=\sum_{k=0}^\infty \left(\frac{2(k+x)}{(k+x)^2+y^2} - \frac{2(k+x+z)}{(k+x+z)^2+y^2}\right)\ge 0 $$ for $x\ge\frac 12, z>0$, which, by Gary's computation of the difference, is equivalent to showing that $$ \sum_{k=0}^\infty \frac{(k+x)^2-y^2}{[(k+x)^2+y^2]^2}\ge 0 $$ for $x\ge\frac 12$ (this is just the derivative of the sum with respect to $z$ with $x+z$ replaced by $x$).
Denoting $w=x+iy$, we see that the last sum can be written as $$ \sum_{k=0}^\infty\Re\frac{1}{(w+k)^2}\,. $$ This series is dominated by $\sum_{k=0}^\infty\frac{1}{(k+\frac 12)^2}$ in the half-plane $\Re w\ge \frac 12$, so it defines a bounded harmonic function there. Thus it suffices to show that it is non-negative on the boundary $x=\Re w=\frac 12$.
However there it can be computed. First, note that $$ \sum_{k=0}^\infty \frac{(k+\frac 12)^2-y^2}{[(k+\frac 12)^2+y^2]^2} =\frac 12\sum_{k=-\infty}^\infty \frac{(k+\frac 12)^2-y^2}{[(k+\frac 12)^2+y^2]^2}\,. $$ Now consider the meromorphic function $$ g(w)=-\frac{\tan(\pi w)}{(w-iy)^2}\,. $$ It has poles at half-integers where the real parts of the residues are exactly what we want to sum, and tends to $0$ like $1/R^2$ on the circles $|w|=R$ with integer $R$, so the sum of all residues must be $0$. Thus, our sum is just minus the real part of the residue of $g$ at $iy$, i.e., $$ \Re\left[\frac d{dw}\tan(\pi w)\right]=\Re\left[\frac \pi {\cos^2(\pi iy)}\right]\,, $$ which is, indeed, some positive number ($\cos(\pi iy)$ is real and not $0$).