Refinement about :$\left(\left(1-x\right)^{-\left(2x\right)}-1\right)\left(\left(x\right)^{-\left(2\left(1-x\right)\right)}-1\right)\geq 1$ Claim :
Let $0.5\leq x<1$ then it seems we have :
$$\left(\left(1-x\right)^{-2x}-\frac{\left(1-x\right)^{2x}\left(x\right)^{2\left(1-x\right)}}{2^{4}\left(x\left(1-x\right)\right)^{3}}\right)\left(x^{-2\left(1-x\right)}-1\right)\geq 1$$

Background :
It's a refinement of :
$$\left(\left(1-x\right)^{-\left(2x\right)}-1\right)\left(\left(x\right)^{-\left(2\left(1-x\right)\right)}-1\right)\geq 1\quad (I)$$
With the same constraint as above wich is an inequality due to Vasile Cirtoaje .
My refinement is based on one single and simple fact :
Let $0.5\leq x<1$ then we have :
$$\frac{\left(1-x\right)^{2x}\left(x\right)^{2\left(1-x\right)}}{2^{4}\left(x\left(1-x\right)\right)^{3}}\geq 1$$
The proof of this fact is not hard taking logarithm and derivative .
Also Vasile Cirtoaje proved the inequality $(I)$ with some tools wich are not sufficient to show the refinement above .Generalising this simple fact it seems work  with Prove that if $a+b =1$, then $\forall n \in \mathbb{N}, a^{(2b)^{n}} + b^{(2a)^{n}} \leq 1$. .


Edit : We have the precious inequality wich simplify the rest on $x\in [0.5,0.75]$ :
$$\left(\left(1-x\right)^{-2x}-\frac{\left(1-x\right)^{2x}\left(x\right)^{2\left(1-x\right)}}{2^{4}\left(x\left(1-x\right)\right)^{3}}\right)\left(x^{-2\left(1-x\right)}-1\right)\geq\frac{\left(1-x\right)^{2x}\left(x\right)^{2\left(1-x\right)}}{2^{4}\left(x\left(1-x\right)\right)^{3}} \geq 1$$
It works also on a larger interval but like this we can use Bernoulli's inequality next.
Edit 2 :
The claim is also :
Let $0.5\leq x \leq 0.75$ then we have :
$$1\geq x^{2\left(1-x\right)}\left(\frac{\left(1-x\right)^{4x}}{2^{4}\left(x\left(1-x\right)\right)^{3}}+1\right)$$
We have also :
Let $0.5\leq x \leq 0.75$ then we have :
$$(1-x)^{4x}\leq \left(2^{2\left(1-x\right)}x\left(1-x\right)^{2}\cdot2\right)^{2}$$
And using Gerber's theorem we have $x\in[0.5,0.75]$:
$$f\left(x\right)=0.5^{2\left(1-x\right)}+2\cdot0.5^{2\left(1-x\right)}\cdot2\left(1-x\right)\left(x-0.5\right)+2\cdot0.5^{2\left(1-x\right)}\cdot\left(2\left(1-x\right)-1\right)\cdot2\left(1-x\right)\cdot\left(x-0.5\right)^{2}+\frac{4}{3}\cdot0.5^{2\left(1-x\right)}\cdot\left(2\left(1-x\right)-1\right)\cdot2\left(1-x\right)\cdot\left(2\left(1-x\right)-2\right)\cdot\left(x-0.5\right)^{3}\geq x^{2(1-x)}$$
Last edit :
We the following inequalities $x\in[0.5,0.55]$
$$h(x)=\left(\frac{2^{-2\cdot\left(1-x\right)}x}{1-2^{\left(0.95-1\right)}\left(\left(1-x\right)2x\right)^{0.95}}\right)\geq x^{2(1-x)}$$
And :
$$\left(\frac{\left(2^{2\left(1-x\right)}x\left(1-x\right)^{2}\cdot2\right)^{2}}{2^{4}\left(x\left(1-x\right)\right)^{3}}+1\right)h(x)\leq 1$$
I think it's not hard using derivatives .



Question :
How to show the claim ?
Thanks.
Reference :
VASILE CIRTOAJE, PROOFS OF THREE OPEN INEQUALITIES WITH POWER-EXPONENTIAL FUNCTIONS, Journal of Nonlinear Sciences and Applications, 4 (2011), no. 2, 130-137
 A: Sketch of a proof:
Denote
$$A = (1 - x)^{2(1 - x)}, \quad B = (1 - x)^{2x - 1},
\quad C = x^{2x}.$$
The desired inequality is written as
$$\left(\frac{A}{(1 - x)^2} - \frac{B}{2^4x(1 - x)^2C}\right)(x^{-2}C - 1) \ge 1.$$
Denote $a = \ln 2$. Let
\begin{align*}
 A_1 &= \frac{p_1x^2 + p_2x + p_3}{(8a^3 - 24a^2
  + 48a - 24) x - 4a^3 - 12}, \\
 B_1 &= \frac{(2a -2)x^2 + (-3a + 2)x + a}{(4a - 2)x^2 + (-4a + 2)x + a}, \\
 C_1 &= \frac{(4a - 2)x^2 + (-4a + 2)x + a}{(2a - 2)x - a + 2},
\end{align*}
where
\begin{align*}
 p_1 &= -4a^4 + 16a^3 -24a^2 + 24a - 24, \\
 p_2 &= 4a^4 - 24a^3 + 48a^2 - 48a + 36, \\
 p_3 &= - a^4 + 8a^3 - 24a^2 + 30a - 24.
\end{align*}
Fact 1: $A \ge A_1 > 0$ for all $x \in [1/2, 1)$.
Fact 2: $B \le B_1$ for all $x \in [1/2, 1)$.
Fact 3: $C \ge C_1 > 0$ for all $x \in [1/2, 1)$.
Fact 4: $x^{-2}C_1 - 1 > 0$ for all $x \in [1/2, 1)$.
By Facts 1-4, it suffices to prove that
$$\left(\frac{A_1}{(1 - x)^2} - \frac{B_1}{2^4x(1 - x)^2C_1}\right)(x^{-2}C_1 - 1) \ge 1$$
which is true (simply a polynomial inequality).
We are done.
