A gamma function inequality I would like to prove $$\frac{\Gamma(n+\frac{1}{2})}{\Gamma(n+1)} \le \frac{1}{\sqrt{n}}$$ for all natural $n \ge 1$. The inequality does seem to be true numerically, but the proof eludes me.
Here's a direct proof:

The gamma function satisfies $\Gamma(n+1) = n \Gamma(n)$,
so the function $g(x) = \ln \Gamma(x)$ satisfies
$g(n+1) = \ln n + g(n)$.
In other words, the line segment between the points $(n,g(n))$
and $(n+1,g(n+1))$ has slope $\ln n$.

Moreover, $g$ is known to be a convex function for $x&gt;0$
(cf. the Bohr–Mollerup theorem),
so the point $(n+1/2, g(n+1/2))$ lies below that line segment:
$g(n+1/2) &lt; g(n+1) - \frac12 \ln n$.
Exponentiating both sides gives $\Gamma(n+1/2) &lt; \Gamma(n+1) / \sqrt{n}$,
as desired.
\begin{eqnarray*}
\Gamma(x+1)&gt;\frac{\frac{(900\gamma^2+73\pi^2)x}{900\gamma}+\frac{73}{100}}{\frac{\pi^2 x}{9\gamma}+1}\quad \textbf{[1]}
\end{eqnarray*}
i want to prove this inquality on $(0,1) is increasing 
i defined it as

i defined it as

\begin{eqnarray*}
w(x)=\frac{\log[\Gamma(x+1)(\frac{\pi^2 x}{9\gamma}+1)]}{\log[\frac{(900\gamma^2+73\pi^2)x}{900\gamma}+\frac{73}{100}]} \quad [2]
\end{eqnarray*}
but [2] not increasing to prove it !
what  it need to get it increasing