Prove or disprove that $\left(f'(x)+g'(x)\right)\geq 2$ It's a  result found with geogebra .

Let us define :
$$f(x)=\left(\Gamma(1-x)\right)^{\frac{1}{1-x}}$$
And :
$$g(x)=\operatorname{W}\left(1-\frac{1}{x-1}\right)$$
Then it seems we have $2\leq x\leq \frac{215}{100}$ :
$$\left(f'(x)+g'(x)\right)\geq 2$$


My attempt :
One can show that :
$$\lim_{x\to 2^+}f''(x)=\infty$$
On the other hand we can show that the limit of $g''(x)$ is finite at $x=2$.We deduce the existence of a constant $\varepsilon$ such that the sum of the derivatives is increasing for $2<x\leq 2+\varepsilon$.


Second attempt :
I found a little trick :
If we show that the antiderivative of the sum of derivative is increasing we win because then the derivative is positive. And the good news we can split in two the problem as ($2\leq x \leq 2+k$)  :
$$h(x)=f(x)-\alpha x$$
$$u(x)=g(x)-\beta x$$
Where $\alpha>\beta$ and $\alpha+\beta=2$ and $h(x)$ and $u(x)$ are increasing
As experiments I have tried $\alpha=1.1$ and $\beta=0.9$ and $k=0.03$
last edit :
I go a little bit further :
To show that the function $u(x)$ is increasing we can substitute :
$$y=\frac{xe^x-2}{xe^x-1}$$
Now find a substitution for $h(x)$ is delicate so I give you a link wich concern also my motivation :
https://mathoverflow.net/questions/12828/inverse-gamma-function


My Motivation:
The reason is simple .I have tried to find a "continuous extension"  of the Gamma function with some nice properties around $x=2$.



How to (dis)prove this?
Thanks in advance for your effort.
 A: Sketch of a proof:
Let
\begin{align*}
 F(x) &= f(x) + \frac{27x^3 - 182x^2 + 394x - 316}{10} - 2x,\\
 G(x) &= g(x) - \frac{27x^3 - 182x^2 + 394x - 316}{10}.
\end{align*}
We have
$f'(x) + g'(x) - 2 = F'(x) + G'(x)$.
It is not difficult to prove that $G'(x) \ge 0$
for all $x \in [2, 215/100]$.
It suffices to prove that
$F'(x) \ge 0$ for all $x \in [2, 215/100]$.
It suffices to prove that, for all $x \in [2, 215/100]$,
$$\Gamma(1 - x)^{\frac{1}{1 - x}}
\left(\frac{\ln \Gamma(1 - x)}{(1 - x)^2} - \frac{\psi(1 - x)}{1 - x}\right) + \frac{81}{10}x^2 - \frac{182}{5}x + \frac{197}{5} \ge 0$$
where $\psi(\cdot)$ is the digamma function defined by $\psi(u) = \frac{\mathrm{d} \ln \Gamma(u)}{\mathrm{d} u} = \frac{\Gamma'(u)}{\Gamma(u)}$.
Fact 1: For all $x \in [2, 215/100]$,
$$\Gamma(1 - x)^{\frac{1}{1 - x}} \ge 
(x - 2) + \frac{10}{11}(1 - \gamma)(x - 2)^2 - \frac{10}{11}(x - 2)^2\ln(x - 2) \ge 0$$
where $\gamma$ is the Euler-Mascheroni constant.
Fact 2: For all $x \in [2, 215/100]$,
$$(x - 2)\left(\frac{\ln \Gamma(1 - x)}{(1 - x)^2} - \frac{\psi(1 - x)}{1 - x}\right)
\ge \frac{1 + \frac67(2 - \gamma)(x - 2) - \frac67(x - 2)\ln(x - 2)}{(1 - x)^2} \ge 0.$$
Using Facts 1-2, it suffices to prove that, for all $x \in [2, 215/100]$,
\begin{align*}
 &\left(1 + \frac{10}{11}(1 - \gamma)(x - 2) - \frac{10}{11}(x - 2)\ln(x - 2)\right)\\
 &\quad \times \left(\frac{1 + \frac67(2 - \gamma)(x - 2) - \frac67(x - 2)\ln(x - 2)}{(1 - x)^2}\right)\\
 &\qquad + \frac{81}{10}x^2 - \frac{182}{5}x + \frac{197}{5} \ge 0.
\end{align*}
