Showing the inequality $f(x)=x^{2(1-x)}+(1-x)^{2x}\leq 1$ for $0My proof is not really natural but I think it works.
We want to show 1:

Let $0<x<1$ such that  then we have :
$$f(x)=x^{2(1-x)}+(1-x)^{2x}\leq 1$$

Case $0<x\leq 0.25$
The proof of this case is due to user Batominovski:
By Bernoulli's inequality we have:
$$(1-x)^{2x}\leq 1-2x^2\quad (1)$$
We have also:
$$x^{2-2x}\leq 2x^2\quad (2)$$
Summing $(1)$ and $(2)$ we get the desired inequality.
Case $0.25\leq x\leq 0.49$:
I shall prove it later but we have :


Let $0.25\leq x\leq 0.49$ then we have :


$$p(x)=2^{2x}(1-x)x^{2}2\geq x^{2(1-x)}$$
And
$$h(x)=\cos^2\Big(x\frac{\pi}{2}\Big)(1+\frac{195}{100}(1-x)(0.5-x)x^2)\geq (1-x)^{2x}\quad (3)$$
With my work we have :
$$f^2(x)\leq \Big(\cos^2\Big(x\frac{\pi}{2}\Big)(1+\frac{195}{100}(1-x)(0.5-x)x^2)\Big)^2+\Big(2^{2x}(1-x)x^{2}2\Big)^2+4^{1.95}(x(1-x))^{2.95}2\leq 1$$
To show it we can use power series see here
Case $0.49\leq x \leq 0.5$
On the domain $[0.49,0.51]$ the function $g(x)=x^{2(1-x)}$ is concave or:
$$g''(x)\leq 0\quad (4)$$
So we have by Jensen's inequality:
$$g(x)+g(1-x)\leq 2g(0.5)=1$$
As $f(x)=f(1-x)$ it's proved for $0.5\leq x<1$
Question:
How to show $(3)$?
Thanks in advance!
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

Actually there is an easier way. Lets search for the maximum of that function. If its outside of $0$ and $1$ than it should be at the boundaries. (But this is not the case)

*

*Differentiate the function.

*Set the derivative equal to zero.

*Solve for $x$.

