How to prove
$$\sqrt{\frac{n-1}{2}} \frac{\Gamma\left[\frac{n-1}{2}\right]}{\Gamma\left[\frac{n}{2}\right]} \gt 1 \quad \forall n \ge 2,n\in \mathbb{N}$$
I plot this function in Mathematica and verify it indeed is greater than $1$. I just know about some basic property of Gamma function.
Any hint? Thanks in advance!
Putting $$ F(n) = {{\sqrt {{{n - 1} \over 2}} \Gamma \left( {{{n - 1} \over 2}} \right)} \over {\Gamma \left( {{n \over 2}} \right)}} $$ then $$ \eqalign{ & F(n)^{\,2} = {{\left( {{{n - 1} \over 2}} \right)\Gamma \left( {{{n - 1} \over 2}} \right)\Gamma \left( {{{n - 1} \over 2}} \right)} \over {\Gamma \left( {{n \over 2}} \right)^{\,2} }} = {{\Gamma \left( {{{n + 1} \over 2}} \right)\Gamma \left( {{{n - 1} \over 2}} \right)} \over {\Gamma \left( {{n \over 2}} \right)^{\,2} }} \cr} $$ and the result follows from the log-convexity of Gamma $$ \ln F(n) = {1 \over 2}\left( {\ln \Gamma \left( {{{n + 1} \over 2}} \right) + \ln \Gamma \left( {{{n - 1} \over 2}} \right)} \right) - \ln \Gamma \left( {{n \over 2}} \right) > 0\quad \left| {\;1 < n \in \mathbb R} \right. $$
At least for "large" values of $n$.
Let $$y=\sqrt{\frac{n-1}{2}}\,\, \frac{\Gamma[\frac{n-1}{2}]}{\Gamma[\frac{n}{2}]} $$ $$\log(y)=\frac 12 \log(n-1)-\frac 12 \log(2) + \log \left(\Gamma \left(\frac{n-1}{2}\right)\right)-\log \left(\Gamma \left(\frac{n}{2}\right)\right)$$ Now, use Stirling approximation $$\log(\Gamma(p))=p (\log (p)-1)+\frac{1}{2} \left(-\log \left({p}\right)+\log (2 \pi )\right)+\frac{1}{12 p}-\frac{1}{360 p^3}+O\left(\frac{1}{p^5}\right)$$ and get (continuing with Taylor expansions) $$\log(y)=\frac{1}{4 n}+\frac{1}{4 n^2}+\frac{5}{24 n^3}+O\left(\frac{1}{n^4}\right)$$ $$y=e^{\log(y)}=1+\frac{1}{4 n}+\frac{9}{32 n^2}+\frac{35}{128 n^3}+O\left(\frac{1}{n^4}\right)$$
