Let $2\leq x\leq 3$ for the function :

$$f(x)=x^{(\Gamma(x+\frac{1}{2}))} (\Gamma(x+\frac{1}{2}))^{(-x)}$$

Then denotes by $x\_{min}$ the abscissa of the minimum .

Prove that :
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$$f(x\_{min})<1$$

I have tried to attack this problem with Taylor's series and we have at $x=3$ :

$$f(x)\simeq 1.04933 + 0.655881 (x - 3) + 2.58005 (x - 3)^2 + 3.62558 (x - 3)^3 + 5.4263 (x - 3)^4 + 7.87526 (x - 3)^5 + O((x - 3)^6)$$

Now the problem is the rest that I cannot cancel easily.

Another approach is the Legendre duplication formula wich states :

$${\displaystyle \Gamma (z)\Gamma \left(z+{\tfrac {1}{2}}\right)=2^{1-2z}\;{\sqrt {\pi }}\;\Gamma (2z).}$$

But I think it complicate the problem more than it solves .

How to prove it ?

Any suggestions is very welcome .