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.)


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 $x\ge2$.
