Problem :
Let :
$$\frac{\Gamma(x+1)}{(x+x^2)}=f(x),x>0$$
Then let $A=\sqrt{\frac{\pi}{4}}-1$ And $y$ be the global minimum over $x\in (0,\infty)$ of $f(x)$ then it seems we have :
$$2+A-A^4-30A^6-103A^8-76A^{10}-23A^{12}-22A^{14}-3A^{16}-\cdots=y\tag{I}$$
I cannot pursue it because of the floating point in computer but I have a Conjecture:
Conjecture :
Starting from $-3A^{16}$ all the coefficient in absolute value are less than $100$.
Update :
A similar series is :
$B=0.4616321944\cdots!$ be the minimum of $\Gamma(x+1),x>0$ then it seems we have :
$$y=1+B+2(B-1)^4+20(B-1)^6+\cdots$$
Why it's interesting:
The related inequality :
$$h(x)=\Gamma(x)\geq f(y)(x+1)=h'(y)(x-y)+h(y)$$
Which says there is no other form of this kind .
I think we need a second point as we have :
$$\arcsin(\int_{0}^{1}t!dt×x)=q(x),\Gamma(x+1)=h(x+1)$$
Then :
$$|q(A+1)-h(A+1)|<4×10^{-5}$$
A strategy :
We have :
$$(x+1)^2\psi(x+2)+(x+1)(\psi(x+2)-2)-1=F(x),F(y-1)=0$$
Where $\psi$ is the digamma function .
Then as the derivative is positive for the truncated sum $I$ we can use Bernstein's polynomials with a faster converging series than the classical Bernstein's polynomials. Then use Bürmann theorem
Question :
How to (dis)prove the conjecture ?
Reference :
Let $$z=2+A-A^4-30A^6-103A^8-76A^{10}-23A^{12}-22A^{14}-3A^{16}.$$ As I understood the conjecture, it states that there exists a sequence $(c_n)_{n=9}^\infty$ of integers of absolute value less that $100$ such that $z+\sum_{n=9}^\infty c_nA^{2n}=y$, that is $y-z=\sum_{n=9}^\infty c_nA^{2n}$. Since $A^2>\frac 1{99}$, it is easy to show that the set of real numbers representable in the form $\sum_{n=9}^\infty c_nA^{2n}$ is the segment $[-C,C]$, where $C=\sum_{n=9}^\infty 99A^{2n}=\frac{99A^{18}}{1-A^2}$. (In fact, to represent the numbers from $[-C,C]$ it suffices to use only sequences $(c_n)_{n=9}^\infty$ with nonpositive or nonnegative members). So the conjecture is equivalent to $|y-z|\le C$. I expect the latter inequality can be checked if we calculate $y$ and $z$ sufficiently precise.