Inequality involving the regularized gamma function Prove that $$Q(x,\ln 2) := \frac{\int_{\ln 2}^{\infty} t^{x-1} e^{-t} dt}{\int_{0}^{\infty} t^{x-1} e^{-t} dt} \geqslant 1 - 2^{-x}$$ for all $x\geqslant 1$. 
($Q$ is the regularized gamma function.)

 A: We have
$$
\frac{\int_{\ln 2}^{\infty} t^{x-1} e^{-t} \,dt}{\int_{0}^{\infty} t^{x-1} e^{-t} \,dt} = \frac{\int_{0}^{\infty} t^{x-1} e^{-t} \,dt - \int_{0}^{\log 2} t^{x-1} e^{-t} \,dt}{\int_{0}^{\infty} t^{x-1} e^{-t} \,dt} = 1 - \frac{\int_{0}^{\log 2} t^{x-1} e^{-t} dt}{\int_{0}^{\infty} t^{x-1} e^{-t} \,dt},
$$
so we need to show that
$$
\frac{\int_{0}^{\log 2} t^{x-1} e^{-t} \,dt}{\int_{0}^{\infty} t^{x-1} e^{-t} \,dt} \leq 2^{-x},
$$
or, equivalently,
$$
2^x \int_{0}^{\log 2} t^{x-1} e^{-t} \,dt \leq \int_{0}^{\infty} t^{x-1} e^{-t} \,dt.
$$
To do this we will show that
$$
2^x \int_{0}^{\log 2} t^{x-1} e^{-t} \,dt \leq \left(\frac{e^a}{e^a-1}\right)^x \int_{0}^{a} t^{x-1} e^{-t} \,dt
\tag{1}
$$
for all $a \geq \log 2$, then let $a \to \infty$.  In fact, we will show that the quantity on the right-hand side of the above inequality is nondecreasing in $a$ when $a > 0$ for fixed $x \geq 1$ (and strictly increasing in $a$ when $a > 0$ for fixed $x > 1$).
To start, define
$$
f_x(a) = \left(\frac{e^a}{e^a-1}\right)^x \int_{0}^{a} t^{x-1} e^{-t} \,dt.
$$
Then
$$
\begin{align}
f_x'(a) &= a^{x-1} e^{-a} \left(\frac{e^a}{e^a-1}\right)^x - x \left(\frac{e^a}{e^a-1}\right)^{x-1} \frac{e^a}{(e^a-1)^2} \int_{0}^{a} t^{x-1} e^{-t} \,dt \\
      &= e^{ax} \left(e^a-1\right)^{-x-1} \left[a^{x-1} \left(1-e^{-a}\right) - x \int_{0}^{a} t^{x-1} e^{-t} \,dt\right].
\end{align}
$$
Since we're only concerned with the sign of the above expression, define
$$
\begin{align}
g_x(a) &= e^{-ax}(e^a - 1)^{x+1} f_x'(a) \\
       &= a^{x-1} \left(1-e^{-a}\right) - x \int_{0}^{a} t^{x-1} e^{-t} \,dt.
\end{align}
$$
If $g_x(a) \geq 0$ for all $a > 0$ then $f_x'(a) \geq 0$ for all $a > 0$, and hence $f_x(a) \geq f_x(\log 2)$ for all $a \geq \log 2$, which is $(1)$.
Well, it will certainly be true that $g_x(a) \geq 0$ for all $a > 0$ if
$$
g_x(0) \geq 0 \hspace{1cm} \text{and} \hspace{1cm} g_x'(a) \geq 0 \,\,\text{ for all }\,\, a \geq 0.
\tag{2}
$$
Indeed, $g_x(0) = 0$, and for $x \geq 1$ we have
$$
g_x'(a) = a^{x-2} e^{-a} (x-1) (e^a - a - 1) \geq 0
$$
since the function $h(a) = e^a - a - 1$ is nondecreasing when $a \geq 0$ and $h(0) = 0$.
By the remarks immediately before $(2)$ this is sufficient to prove $(1)$, from which the result follows.
A: Here is a proof for $x\ge2$:
$$
\begin{align}
\int_0^{\log(2)}t^{x-1}e^{-t}\mathrm{d}t
&\le\int_0^{\log(2)}t^{x-1}\mathrm{d}t\\
&=\frac1x\log(2)^x\tag{1}
\end{align}
$$
Thus, we get that
$$
\frac{\int_0^{\log(2)}t^{x-1}e^{-t}\mathrm{d}t}{\int_0^\infty t^{x-1}e^{-t}\mathrm{d}t}
\le\frac{\log(2)^x}{\Gamma(x+1)}\tag{2}
$$
For $x\ge2$,
$$
\frac{(2\log(2))^x}{\Gamma(x+1)}\le1\tag{3}
$$
Once we show $(3)$, the result follows because
$$
\begin{align}
\frac{\int_{\log(2)}^\infty t^{x-1}e^{-t}\mathrm{d}t}{\int_0^\infty t^{x-1}e^{-t}\mathrm{d}t}
&=1-\frac{\int_0^{\log(2)}t^{x-1}e^{-t}\mathrm{d}t}{\int_0^\infty t^{x-1}e^{-t}\mathrm{d}t}\\
&\ge1-\frac{\log(2)^x}{\Gamma(x+1)}\\
&\ge1-2^{-x}\tag{4}
\end{align}
$$

Inequality $(3)$ is equivalent to
$$
\log(\Gamma(x+1))\ge x(\log(2\log(2)))\tag{5}
$$
Note that $(5)$ holds at $x=2$ since $\log(2)>2\log(2\log(2))$ follows from $\log(2)<\sqrt{1/2}$.

Since $\Gamma$ is log-convex and for $x\ge2$, $\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x+1))\ge\frac32-\gamma>\log(2\log(2))$.
Thus, $(5)$, and therefore $(3)$, hold for $x\ge2$.
