# How to show that $\frac{\sqrt{n-1}\Gamma((n-1)/2)}{\sqrt{2}\Gamma(n/2)}>1$.

Show that $$\frac{\sqrt{n-1}\Gamma((n-1)/2)}{\sqrt{2}\Gamma(n/2)}>1$$

I tried to solve it using Taylor series expansion.

In this answer, it is shown that $\Gamma(x)$ is log-convex. Therefore, \begin{align} \Gamma\left(\frac n2\right) &\le\Gamma\left(\frac{n-1}2\right)^{1/2}\Gamma\left(\frac{n+1}2\right)^{1/2}\\[6pt] &=\sqrt{\frac{n-1}2}\ \Gamma\left(\frac{n-1}2\right) \end{align} since $\Gamma(x+1)=x\Gamma(x)$.
A log-convex function is a function whose logarithm is convex. Thus, if $f$ is log-convex, we have $$\log\circ f\left(\frac{x+y}2\right)\le\frac{\log\circ f(x)+\log\circ f(y)}2$$ removing the $\log$s by applying $\exp$, which is monotonically increasing, we get $$f\left(\frac{x+y}2\right)\le f(x)^{1/2}f(y)^{1/2}$$