Prove that $f(x_0)>\frac{2}{3}$ It's a problem found with the help of Geogebra.
Let $0<x$ be a real number then define the function:
$$f(x)=\Big(\frac{x}{x+1}\Big)^{\Gamma(x)}$$
Then let $x_0$ be the maximum of the function on $(0,\infty)$ and  then prove that:
$$f(x_0)>\frac{2}{3}$$
See here to compare
Well to solve it I have tried logically the use of derivative we have:
$$f'(x)=\Big(\frac{x}{x+1}\Big)^{\Gamma(x)} \Bigg(\frac{(x + 1) \Big(\frac{1}{(x + 1)} - \frac{x}{(x + 1)^2}\Big) Γ(x)}{x} + \log\Big(\frac{x}{x + 1}\Big) Γ(x) \psi^{(0)} (x)\Bigg)$$
Where we have the $n^{th}$ derivative of the digamma function.
I think that this derivative is not really useful only theoretically, but we can use the Newton's method numerically .
I have tried some inequality on the this wiki page notably an inquality due to  Kečkić and Vasić without success.
On the other hand the problem with Taylor series is : we get a lot of constant as Euler-Mascheroni constant wich needs to be evaluate with an series or something like that. So it's a little bit make problem on another problem.
Maybe spline cubic is the way I don't know...
Finally taking the logarithm on both side the derivative is a little bit less tedious.See here
Well if you have an issue thanks in advance ...
 A: Hint
Try to expand $f$ at the first order around $2$ based on
$g(x) = \frac{x}{x+1} = \frac{2}{3}(1+h/6) +o(h^2)$ where $x=2+h$ and $\Gamma(2+h)=1+(1-\gamma)h+o(h^2)$ where $\gamma$ is the Euler Mascheroni constant.
Therefore
$$\begin{aligned}
\ln f(2+h) &= (1+(1-\gamma)h+o(h^2))(\ln(2/3) + h/6 + o(h^2))\\
&=\ln(2/3) + ((1-\gamma)\ln(2/3) + 1/6)h +o(h^2)
\end{aligned}
$$ proving that $f$ takes around $2$ values larger than $2/3$ as $(1-\gamma)\ln(2/3) + 1/6 \neq 0$.
A: Almost the same as in comments and answers.
Since $x>0$, maximizing
$$f(x)=\Big(\frac{x}{x+1}\Big)^{\Gamma(x)}$$ is the same as maximizing
$$g(x)=\Gamma(x) \log\Big(\frac{x}{x+1}\Big)$$ for which
$$\frac{g'(x)}{g(x)}=\Gamma (x) \left(\frac{1}{x(x+1)}+\log \left(\frac{x}{x+1}\right) \psi
   (x)\right)$$ and, as already said, the quantity inside parentheses cancels close to $x=2$. Using one single iteration of Newton, Halley, Householder and higher order iterative methods of the same class, we obtain totally explicit expressions of $x_0$ corresponding to the maximum of $f(x)$. Since the formulae can be quite long, only their decimal representation will be given as a function of $n$ (the order of the method).
$$\left(
\begin{array}{ccc}
n & x_0^{(n)} & \text{method} \\
 2 &  1.985579580 & \text{Newton}\\
 3 &  1.985734229 & \text{Halley}\\
 4 &  1.985733904 & \text{Householder}\\
\cdots & \cdots & \text{no name}\\
\infty & 1.985733904 &
\end{array}
\right)$$ So, $$x_0^{(2)}=2+\frac{36 (\gamma -1) \log \left(\frac{3}{2}\right)-66}{35+6 \gamma +6 \left(\pi^2-6\right) \log \left(\frac{3}{2}\right)}$$ seems to be a sufficient approximation.
$$f(x_0^{(2)})=\Big(\frac{x_0^{(2)}}{x_0^{(2)}+1}\Big)^{\Gamma(x_0^{(2)})}\approx 0.6666893243$$  Notice that a full optimization gives a maximum of $0.6666893270$ for $x=1.985733903$.
