<s>[INST] 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.
 [/INST] 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.
$$ \frac{1}{ \sqrt{n} } > \frac{ \Gamma(n + 1/2) }{ \Gamma(n+1) } > \frac{1}{ \sqrt{ n + 1/2} }.$$
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