Finding a closed form to a minimum of a function It's a try to find a closed form to the minimum of the function :
Let $0<x<1$ then define :
$$g(x)=x^{2(1-x)}+(1-x)^{2x}$$
Denotes $x_0$ the abscissa of the minimum .
Miraculously using Slater's inequality for convex function I have found that :
Define $f(x)=x^{2(1-x)}$ then :
$$\lim_{x\to x_0}\Bigg(0.5+\frac{(x-1)f'(x)-xf'(1-x)}{f'(x)+f'(1-x)}\Bigg)=0$$
And by the definition of the derivative :
$$\lim_{x\to x_0}g'(x)=0$$
For the first limit see here to compare with the second limit see here
My question :
With these two equations can we hope to find a nice closed form ?
Any helps is greatly appreciated
Thanks in advance because it's a hard nut .
Little update
Well I have got a down-vote there is no mistake if we  use natural logarithm for the first limit .
 A: Partial results since I am still working the problem.
Using the Inverse Symbolic Calculator, the point $x_*$ where the derivative cancels seems to be very close to
$$x_*=10 \,(\gamma \, K_1(1)){}^{\Gamma \left(\frac{1}{4}\right)}$$ which numerically is
$$x_*=0.216453828\qquad \implies g(x_*)=0.99066450008687554$$
while the exact results are
$$x_*=0.216453839\qquad \implies g(x_*)=0.99066450008687550$$
A much better approximation seems to be
$$x_*=\frac{97+800 e-307 e^2}{-263-205 e+113 e^2}$$ which is in an absolute error of $2.87 \times 10^{-19}$; for this value, $g(x_*)$ is in an absolute error of $2.76 \times 10^{-28}$.
A: Because $x_0$ the abscissa of the minimum we know that $g'(x_0)=0$ then you can calculate the function $g'(x)$:
$$g'(x) = \bigr(x^{2(1−x)} \bigl)'+\bigr((1−x)^{2x}\bigl)' = \bigr(e^{\ln{(x^{2(1−x)})}} \bigl)'+\bigr(e^{\ln{((1−x)^{2x})}}\bigl)' = \bigr(e^{{2(1−x)} \ln{(x)}} \bigl)'+\bigr(e^{{2x} \ln{(1−x)}}\bigl)' = 
\bigl[ \bigr(e^{{2(1−x)} \ln{(x)}} \bigl)\bigr({{2(1−x)} \ln{(x)}} \bigl)' \bigl]+\bigl[ \bigr(e^{{2x} \ln{(1−x)}}\bigl)({2x} \ln{(1−x)})' \bigr] = $$
$$ = \bigl[x^{2(1−x)}\bigl(2(1-x){1\over x}-2\ln(x) \bigr) \bigr]+\bigl[(1-x)^{2x}\bigl(2\ln(1-x)+2{x\over (1-x)}\bigr)\bigr] =
2\bigl[x^{2(1−x)}({1\over x}-1-\ln(x))+(1-x)^{2x}(\ln(1-x)+{x\over (1-x)})\bigr]$$
Now because $g'(x_0)=0$:
$$
2\bigl[x_0^{2(1−x_0)}({1\over x_0}-1-\ln(x_0))+(1-x_0)^{2x_0}(\ln(1-x_0)+{x_0\over (1-x_0)})\bigr]=0$$
so we can get:
$$x_0^{2(1−x_0)}({1\over x_0}-1-\ln(x_0))=-(1-x_0)^{2x_0}(\ln(1-x_0)+{x_0\over (1-x_0)})$$
$$$$
The solutions are: $x_0 = 0.216... \ , \ 0.783... \ \ and \ \ 0.5$, but you can easily find $x$ such as $g(x)<g(0.5)$ so $0.5$ is the max point and $x_0 = 0.216... \ , \ 0.783...$ are the minimum points.
You can sea the solution here: https://www.wolframalpha.com/input/?i=-2+x%5E%281+-+2+x%29+%28x+%2B+x+log%28x%29+-+1%29+-+2+%281+-+x%29%5E%282+x+-+1%29+%28x+%2B+%28x+-+1%29+log%281+-+x%29%29+%3D0.
