Show that $2^{ax}\frac{\Gamma((a+1)x)}{\Gamma(x)}$ is an increasing function I would like to show that the following function 
\begin{align}
f_a(x)=2^{ax}\frac{\Gamma((a+1)x)}{\Gamma(x)}
\end{align}
is an increasing function in $x$ for $x \ge 0$  for any fixed $a>0$.
I did some simulations but not sure how to show a proof for this. I wanted to point out that, from simulation, it seems that $\frac{\Gamma((a+1)x)}{\Gamma(x)}$ can be decreasing for values  $x=0$. 
We can attempt this by showing that the derivative  of a logarithm of $f_a(x)$ is positive. Let 
\begin{align}
g_a(x)= \log (f_a(x))
\end{align}
(here log is base e) and the derivative of $g_a(x)$ is given by
\begin{align}
\frac{d}{dx}  g_a(x)&=  \frac{d}{dx} \left( ax \log(2)+  \log (\Gamma( (a+1)x))- \log (\Gamma( x)) \right)\\
&=a \log(2) + (a+1)\psi((a+1)x)-\psi(x)
\end{align} 
where $\psi(x)$ is a digamma function. 
Now it remains to show the following inequality for the difference of digamma functions
\begin{align}
 (a+1)\psi((a+1)x)-\psi(x) \ge - a \log(2) .
\end{align}
 A: We have that
$$
r_a(x) = {{\Gamma \left( {\left( {a + 1} \right)x} \right)} \over {\Gamma \left( x \right)}}
 = {{\Gamma \left( {x + a\,x} \right)} \over {\Gamma \left( x \right)}} = x^{\,\overline {\,a\,x\,} } 
$$
where $x^{\,\overline {\,y\,} } $ denotes the Rising Factorial.
Now, the Rising Factorial is defined (for $x$ and $y$ real and also complex) as
$$
h(x,y) = x^{\,\overline {\,y\,\,} }  = {{\Gamma \left( {x + y} \right)} \over {\Gamma \left( x \right)}} = \prod\nolimits_{\;k\, = \,\,0}^{\,y} {\left( {x + k} \right)} 
$$
where the last term denotes the Indefinite Product, computed for $k$ ranging between the indicated bounds.
So we can write $f_a(x)$ as
$$ \bbox[lightyellow] {  
f_{\,a} (x) = 2^{\,a\,x} {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}
 = \prod\nolimits_{\;k\, = \,\,0}^{\,a\,x} {2\left( {x + k} \right)}  = 2^{\,a\,x} x^{\,a\,x} \prod\nolimits_{\;k\, = \,\,0}^{\,a\,x} {\left( {1 + k/x} \right)} 
} \tag{1}$$
Concerning the derivative of $r_a(x)$, since
$$
\left\{ \matrix{
  {\partial  \over {\partial x}}h(x,y) = {{\Gamma \left( {x + y} \right)} \over {\Gamma \left( x \right)}}\left( {\psi \left( {x + y} \right)
 - \psi \left( x \right)} \right) \hfill \cr 
  {\partial  \over {\partial y}}h(x,y) = {{\Gamma \left( {x + y} \right)} \over {\Gamma \left( x \right)}}\psi \left( {x + y} \right) \hfill \cr}  \right.
$$
then, as you already found
$$
\eqalign{
  & {d \over {dx}}r_a(x) = {\partial  \over {\partial x}}h(x,y) + {\partial  \over {\partial y}}h(x,y){d \over {dx}}y =   \cr 
  &  = {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}\left( {\left( {a + 1} \right)\psi \left( {x + ax} \right)
 - \psi \left( x \right)} \right) =   \cr 
  &  = {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}\left( {a\psi \left( {x + ax} \right) + \left( {\psi \left( {x + ax} \right)
 - \psi \left( x \right)} \right)} \right) \cr} 
$$
where, for $0<x$ (and $0<a$) $\Gamma(x+ax)/\Gamma(x)$ is clearly positive.
However, while  $\psi(x+ax)-\psi(x)$ is also positive since $\psi(x)$ is increasing in that range,  $a\psi(a+ax)$ introduces a negative term for lower $x$.
To determine the limit of $r_a'(x)$ as $x \to 0^+$, let's consider the series development of 
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  \ln \Gamma (cx) = \ln \left( {{1 \over {cx}}} \right) - \gamma cx + O\left( {x^{\,2} } \right) \hfill \cr 
  \psi (cx) =  - {1 \over {cx}} - \gamma  + {{\pi ^{\,2} } \over 6}cx + O\left( {x^{\,2} } \right) =  \hfill \cr 
   =  - {1 \over {cx}} - \gamma  + \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {1 \over {k + 1 + cx}}} \right)}  \hfill \cr}  \right.
} \tag{2}$$
Therefore
$$
\eqalign{
  & \mathop {\lim }\limits_{x\; \to \;0^{\, + } } {d \over {dx}}r_a(x) = \mathop {\lim }\limits_{x\; \to \;0^{\, + } } {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}\mathop {\lim }\limits_{x\; \to \;0^{\, + } } \left( {\left( {a + 1} \right)\psi \left( {x + ax} \right) - \psi \left( x \right)} \right) =   \cr 
  &  = {1 \over {a + 1}}\left( { - \gamma a} \right) \cr} 
$$
which is negative for $0<a$.
Concerning $f_a(x)$ instead
$$
\eqalign{
  & {d \over {dx}}f_{\,a} (x) = \;2^{\,a\,x} a\ln 2r_{\,a} (x) + 2^{\,a\,x} {d \over {dx}}r_{\,a} (x) =   \cr 
  &  = 2^{\,a\,x} r_{\,a} (x)\left( {a\ln 2 + {d \over {dx}}\ln \left( {r_{\,a} (x)} \right)} \right) =   \cr 
  &  = 2^{\,a\,x} r_{\,a} (x)\left( {a\ln 2 + {d \over {dx}}\ln \Gamma (x + ax) - {d \over {dx}}\ln \Gamma (x)} \right) =   \cr 
  &  = 2^{\,a\,x} r_{\,a} (x)\left( {a\ln 2 + \left( {a + 1} \right)\psi (x + ax) - \psi (x)} \right) \cr} 
$$
(which is the equation you already found)   
and 
$$ \bbox[lightyellow] {  
\eqalign{
  & \mathop {\lim }\limits_{x\; \to \;0^{\, + } } f_{\,a} '(x) = 1\left( {\left( {a\ln 2} \right){1 \over {a + 1}} - {{\gamma a} \over {a + 1}}} \right) =   \cr 
  &  = {a \over {a + 1}}\left( {\ln 2 - \gamma } \right) = {a \over {a + 1}}0.1159 \cdots  \cr} 
} \tag{3}$$
Proceeding with the development of the derivative above
$$ \bbox[lightyellow] {  
\eqalign{
  & {d \over {dx}}f_{\,a} (x)\;\mathop /\limits_{} \;\left( {2^{\,a\,x} r_{\,a} (x)} \right) =   \cr 
  &  = a\ln 2 + \left( {a + 1} \right)\psi (x + ax) - \psi (x) =   \cr 
  &  = a\ln 2 + \left( {a + 1} \right)\left( { - {1 \over {\left( {a + 1} \right)x}} - \gamma  + \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}}
 - {1 \over {k + 1 + \left( {a + 1} \right)x}}} \right)} } \right) - \left( { - {1 \over x} - \gamma  + \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}}
 - {1 \over {k + 1 + x}}} \right)} } \right) =   \cr 
  &  = \left( {a\ln 2 - {1 \over x} - \left( {a + 1} \right)\gamma  + {1 \over x} + \gamma } \right)
 + \left( {a + 1} \right)\sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {1 \over {k + 1 + \left( {a + 1} \right)x}}} \right)}
  - \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {1 \over {k + 1 + x}}} \right)}  =   \cr 
  &  = a\left( {\ln 2 - \gamma } \right) + \sum\limits_{0\, \le \,k} {\left( {{a \over {k + 1}} + {1 \over {k + 1 + x}}
 - {{\left( {a + 1} \right)} \over {k + 1 + \left( {a + 1} \right)x}}} \right)}  =   \cr 
  &  = a\left( {\ln 2 - \gamma } \right) + a\sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}}
 - {{k + 1} \over {\left( {k + 1 + x} \right)\left( {k + 1 + \left( {a + 1} \right)x} \right)}}} \right)}  \cr} 
} \tag{4}$$
and
$$ \bbox[lightyellow] {  
\eqalign{
  & 0 \le {1 \over {k + 1}} - {{k + 1} \over {\left( {k + 1 + x} \right)\left( {k + 1 + \left( {a + 1} \right)x} \right)}} =   \cr 
  &  = {1 \over {k + 1}} - {1 \over {\left( {1 + x/\left( {k + 1} \right)} \right)\left( {k + 1 + \left( {a + 1} \right)x} \right)}}\quad \left| {\;0 \le x,k} \right. \cr} 
} \tag{5}$$
therefore your thesis is demonstrated.
A: At least for large $x$, considering $$f_a(x)=2^{ax}\frac{\Gamma((a+1)x)}{\Gamma(x)}$$ take logarithms
$$\log(f_a(x) )=a x\log(2)+\log (\Gamma ((a+1) x))-\log (\Gamma ( x))$$ Now, use Stirling approximation for $\log(\Gamma(p))$ and continue with Taylor expansion for large $x$ and get
$$\log(f_a(x) )=x \left(a \log \left(\frac{2x}{e}\right)+(a+1) \log
   (a+1)\right)+\frac{1}{2} \log \left(\frac{1}{a+1}\right)-\frac{a}{12 (a+1)
   x}+O\left(\frac{1}{x^4}\right)$$
A: Sketch of proof:
The ratio $G^a(x):=\Gamma(x+ax)/\Gamma(x)$ for $a>0$ is a function that has a single minimum for $x>0$, and the location of that minimum in the limit that $a \to 0$ (but not exactly zero) is the root of the equation $x\,\psi^{\,'}(x) + \psi(x)=0.$ Call this value $x_0$ which has the numerical value of approximately 0.216099. For $a>>0$, the minimum of $G^a(x)$ occurs for some $0<x^*<x_0.$ Since we want an inequality valid for $x \ge 0$ it is natural to look at $G_a(x)$ for $x \sim 0.$ A linear approximation is 
$G^a_1(x) = 1/(1+a) - a/(1+a)\gamma \, x,$ where $\gamma$ is Euler's constant.  The linear approximation will undershoot the value $G^a(x^*)$ which is what will ultimately result in an inequality.  To counteract the falling trend from 0 to $x^*$ we seek to find $y$ such that
$$\frac{d}{dx} y^x\, G^a(x) |_{x=0} = 0 \text{ or its approx } \frac{d}{dx} y^x\, G^a_1(x) |_{x=0} = 0.$$  This is posed as a question involving a derivative because we want to find the best $y.$ Doing the last calculation it is found that $\log{y}=a\, \gamma.$  However $2^{a\,x}> (e^\gamma)^{a\,x}$ since $e^\gamma \sim 1.78$.  Thus the proposed inequality is true and can be improved with 2 replaced by $e^\gamma$.  To make this argument rigorous will take some work, but it may only need an appeal to Bohr-Mollerup.
