Stronger statement for $x^{2(1-x)}+(1-x)^{2x}\leq 1$ It's an attempt to solve the problem cited here https://www.isr-publications.com/jnsa/articles-1795-a-stronger-inequality-of-cirtoajes-one-with-power-exponential-functions
Problem :
Let $0<x<\frac{1}{2}$ and  $0<r<\frac{1}{2}$ then we have :
$$x^{2(1-x)}+(1-x)^{2(x)}+r((1-x)x(2x-1))^2\leq 1$$
I have already a proof for the classic case 1.I try to show it in the case $r=0.25$.
I give you some details of my attempt :

Let $0.38<x<0.5$ such that  then we have :
$$f(x)=x^{2(1-x)}+(1-x)^{2x}\leq q(x)=(1-x)^{2x}+2^{2x+1}(1-x)x^2\leq 1$$

The Lhs is equivalent to :
$$x^{2(1-x)}\leq h(x)=2^{2x+1}(1-x)x^2$$
Or :
$$\ln\Big(x^{2(1-x)}\Big)\leq \ln\Big(2^{2x+1}(1-x)x^2\Big)$$
Making the difference of these logarithm and introducing the function :
$$g(x)=\ln\Big(x^{2(1-x)}\Big)-\ln\Big(2^{2x+1}(1-x)x^2\Big)$$
The derivative is not hard to manipulate and we see that it's positive and $x=0.5$ is an extrema .The conclusion is :
$$g(x)\leq g(0.5)=0$$
And we are done with the LHS.
For the Rhs I use one of the lemma (7.1) due to Vasile Cirtoaje we have :
$$(1-x)^{2x}\leq p(x)=1-4(1-x)x^{2}-2(1-x)x(1-2 x)\ln(1-x)$$
So we have :
$$q(x)\leq p(x)+h(x)$$
We want to show that :
$$p(x)+h(x)+0.25\left(x\left(1-x\right)\left(2x-1\right)\right)^{2}\leq 1$$
Wich is equivalent to :
$$-2(x-1)x((4^x-2)x+(2x-1)\ln(1-x))+0.25\left(x\left(1-x\right)\left(2x-1\right)\right)^{2}\leq 0$$
It's not hard so I omitt here the proof of this fact .
We are done .
Edit :
Using Bernoulli's inequality we need to show :
$$1-2x^2+x^{2(1-x)}+0.25(x(1-x)(2x-1))^2\leq 1$$
Or :
$$h(x)=x^{2(1-x)-2}+0.25((1-x)(2x-1))^2-2\leq 0$$
The derivative is :
$$h'(x)=4(x-1)(x-0.75)(x-0.5)-2x^{-2x}(\ln(x)+1)$$
The derivative is positive on $(0,0.25)$  wich is obvious because we have :
$$h'(x)\geq 4(x-1)(x-0.75)(x-0.5)-2(\ln(x)+1)>0$$
Remains to evaluate the function at a single point .We can choose $x=0.2$.
We are done again .
Question :
How to complete the proof on $(0.2,0.38)$?
1 Vasile Cirtoaje, "Proofs of three open inequalities with power-exponential functions",
The Journal of Nonlinear Sciences and its Applications (2011), Volume: 4, Issue: 2, page 130-137.
https://eudml.org/doc/223938
 A: Probably one more of my simplistic solutions.
$$f(x)=x^{2(1-x)}+(1-x)^{2x}$$ to be considered $\in \left(0,\frac 12\right)$.
The derivatives do not make any problem. At the bounds
$$f(0)=1 \qquad f'(0)=0\qquad f''(0)=-2$$
$$f\left(\frac 12\right)=1\qquad f'\left(\frac 12\right)=0\qquad f''\left(\frac 12\right)=-4 \left(2-\log ^2(2)-2\log (2)\right) <0$$
So, the bounds correspond to maximum points.
Concerning the minimum : if we perform one single iteration of high-order iterative methods (order $n$) for solving $f'(x)=0$ with $x_0=\frac 14$, we have explicit approximations of the minimum. These are given below
$$\left(
\begin{array}{cccc}
 n & x_{\text{min}}& f_{\text{min}} & \text{method}\\
 2 & 0.213073 & 0.990668 & \text{Newton} \\
 3 & 0.217393 & 0.990665 & \text{Halley} \\
 4 & 0.216372 & 0.990665 & \text{Householder} \\
 5 & 0.216472 & 0.990665 & \text{no name} \\
 6 & 0.216452 & 0.990665 & \text{no name} \\
 7 & 0.216454 & 0.990665 & \text{no name}
\end{array}
\right)$$
A: With the same proof as above for the first part we have on $(0.2,0.38)$ :
$$\frac{1}{4}\left(x\left(1-x\right)\left(2x-1\right)\right)^{2}+1-4\left(1-x\right)^{2}x-2x\left(1-x\right)\left(2x-1\right)\ln\left(x\right)+2^{\left(2x\right)^{0.95}}\left(1-x\right)x\cdot\left(2x\right)^{1.05}\leq 1$$
Obviously it's greater than the Cirtoaje'inequality and it's not hard using derivative as above again.
