A gamma function inequality

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 $x>0$ (cf. the Bohr–Mollerup theorem), so the point $(n+1/2, g(n+1/2))$ lies below that line segment: $g(n+1/2) < g(n+1) - \frac12 \ln n$. Exponentiating both sides gives $\Gamma(n+1/2) < \Gamma(n+1) / \sqrt{n}$, as desired.

For a completely elementary proof let

$$I_n= \int_0^{ \pi/2} \sin^n x \textrm{ d}x $$

then integration by parts gives

$$I_n = \frac{n-1}{n} I_{n-2} . \quad (1) $$

From which we get

$$I_{2n}= \frac{2n-1}{2n} \frac{2n-3}{2n-2} \cdots \frac{1}{2} \frac{\pi}{2} \quad (2) $$

and

$$I_{2n+1}= \frac{2n}{2n+1} \frac{2n-2}{2n-1} \cdots \frac{2}{3} . \quad (3) $$

Since $ \sin^{2n+1} x < \sin^{2n} x $ for $ x \in (0,\pi/2) $ we have

$$I_{2n+1}< I_{2n} . \quad (4) $$

Also, from $(1)$ we have $(2n+1)I_{2n+1} = 2nI_{2n-1} > 2nI_{2n}$ since $I_{2n-1} >I_{2n}.$

Therefore using this and $(4)$

$$ \frac{2n+1}{2n} > \frac{ I_{2n} }{ I_{2n+1} } > 1 . \quad (5) $$

But from $(2)$ and $(3)$

$$ \frac{ I_{2n} }{ I_{2n+1} } = \frac{2n+1}{2} \frac{ (2n!)^2 }{ 4^{2n} (n!)^4 } \pi .$$

Putting this in $(5)$ gives

$$ \frac{1}{ \sqrt{n} } > \frac{1}{4^n} { 2n \choose n } \sqrt{ \pi } > \frac{1}{ \sqrt{ n + 1/2} }.$$

i.e.

$$ \frac{1}{ \sqrt{n} } > \frac{ \Gamma(n + 1/2) }{ \Gamma(n+1) } > \frac{1}{ \sqrt{ n + 1/2} }.$$

This follows directly from Gautschi's inequality, valid for $0 < s < 1$ and $x > 0$:

$$x^{1-s} < \frac{\Gamma(x+1)}{\Gamma(x+s)} < (x+1)^{1-s}.$$

As a side note, the link is to NIST's new (as of this year) Digital Library of Mathematical Functions. The DLMF is an update of the classic Abramowitz and Stegun text, Handbook of Mathematical Functions.

To elaborate on Shai Covo's answer:

$$2n\left[{2n\choose n} 4^{-n}\right]^2 = {1\over2}\,{3\over2}\,{3\over4}\,{5\over4}\cdots {{2n-1}\over{2n-2}}\,{{2n-1}\over{2n}} ={1\over 2}\prod_{j=2}^n\left(1+{1\over4j(j-1)}\right). $$

By Wallis's formula, the middle expression converges to $2\over\pi$. The right hand expression shows that it is strictly increasing. Therefore, multiplying by $\pi/2$ and taking square roots shows that
$$\sqrt{n\pi}{2n\choose n} 4^{-n}\uparrow 1.$$

Since $$ \Gamma \bigg(n + \frac{1}{2}\bigg) = \frac{{(2n)!}}{{4^n n!}}\sqrt \pi $$ (see e.g. here), we have $$ \frac{{\Gamma (n + \frac{1}{2})}}{{\Gamma (n+1)}} = \frac{{{2n \choose n}}}{{4^n }}\sqrt \pi . $$ This might be very useful, but I have to check.