Claim
We want to show [1]:
Let $0.36\leq x\leq 0.5$ and $1\leq k\leq n$ two naturals numbers with $n\geq 10^{10}$ then we have : $$P(k)=(1-x)^{(2x)^{1+\frac{k}{n}}}+x^{(2(1-x))^{1+\frac{k}{n}}}\leq 1\quad (I)$$
Sketch\Partial (of) proof .
We use a form of the Young's inequality or weighted Am-Gm :
Let $a,b>0$ and $0<v<1$ then we have :
$$av+b(1-v)\geq a^vb^{1-v}$$
Taking account of this theorem and putting :
$a=(x)^{(2(1-x))^{1+\frac{k}{n}}}$$\quad$$b=1$$\quad$$v=(2(1-x))^{\frac{-1}{n}}$ we get :
$$(x)^{(2(1-x))^{1+\frac{k-1}{n}}}\leq (x)^{(2(1-x))^{1+\frac{k}{n}}}(2(1-x))^{\frac{-1}{n}}+1-(2(1-x))^{\frac{-1}{n}}$$
Now the idea is to show :
Let $$(1-x)^{(2x)^{1+\frac{k-1}{n}}}\leq 1-\Big((x)^{(2(1-x))^{1+\frac{k}{n}}}(2(1-x))^{\frac{-1}{n}}+1-(2(1-x))^{\frac{-1}{n}}\Big)$$
Or :
$$(1-x)^{(2x)^{1+\frac{k-1}{n}}}\leq (2(1-x))^{\frac{-1}{n}}(1-(x)^{(2(1-x))^{1+\frac{k}{n}}})$$
Or: $$(1-x)^{(2x)^{1+\frac{k-1}{n}}+\frac{1}{n}}+2^{\frac{-1}{n}}(x)^{(2(1-x))^{1+\frac{k}{n}}}\leq 2^{\frac{-1}{n}}\quad (0)$$
Now we want to show that :
$$2^{\frac{-1}{n}}(x)^{(2(1-x))^{1+\frac{k}{n}}}\leq (x)^{(2(1-x))^{1+\frac{k+1}{n}}}\quad(1)$$
For that we need a lemma :
Lemma :
Let $0< x\leq 0.5$ and $1\leq k\leq n$ two naturals numbers then we have :
$$f(k)=\ln(x)((2(1-x))^{1+\frac{k-1}{n}}-(2(1-x))^{1+\frac{k}{n}})\leq f(k+1)=\ln(x)((2(1-x))^{1+\frac{k}{n}}-(2(1-x))^{1+\frac{k+1}{n}})$$
It's true because it's equivalent to :
$$(2(1-x))^{\frac{k-1}{n}}((2(1-x))^{\frac{1}{n}}-1)^2\geq 0$$
So we have :
$$f(n)\geq f(k)$$
We can show also that on $x\in[0.36,0.5]$ :
$$f(k)\leq 0$$
Because we have $x\in[0.36,0.5]$ :
$$p(x)=-\ln(x)((2(1-x))^{1+\frac{k-1}{n}})\geq 0$$
Is decreasing and :
$$t(x)=((2(1-x))^{\frac{1}{n}}-1)\geq 0$$
As the product of two positive decreasing function is a decreasing function remains to calculate the value around $x\simeq 0.36$ wich is equivalent to $(1)$
So we have from $(1)$ and $(0)$ we need to show :
$$ (x)^{(2(1-x))^{1+\frac{k+1}{n}}}+(1-x)^{(2x)^{1+\frac{k-1}{n}}+\frac{1}{n}}\leq 2^{\frac{-1}{n}}$$
But we have $x \in[0.36,0.5]$:
$$(1-x)^{(2x)^{1+\frac{k-1}{n}}+\frac{1}{n}}<(1-x)^{(2x)^{1+\frac{k+1}{n}}}$$
It's true because it's equivalent to :
$$(2x)^{1+\frac{k-1}{n}}-(2x)^{1+\frac{k+1}{n}}\geq 0\geq \frac{-1}{n}$$
So we need to prove $x\in[0.36,0.5-\varepsilon_n]$ with $\varepsilon_n>0$:
$$Q(k+1)=(x)^{(2(1-x))^{1+\frac{k+1}{n}}}+(1-x)^{(2x)^{1+\frac{k+1}{n}}}\leq 2^{\frac{-1}{n}}\quad (2)$$
So we have in fact : $$Q(k+1)\Longrightarrow P(k-1)$$
Now the idea would be to put $k\longrightarrow \infty$ and $n\longrightarrow \infty$ such that $\frac{k}{n}=constant$ but I'm not sure if it's works!So I'm stuck here...
Update 11/11/2020
We have on $x\in[0.15,0.5]$ and $1\leq k\leq n$ two naturals numbers with $n\geq 10^{10}$:
$$P(k)=(1-x)^{(2x)^{1+\frac{k}{n}}}+x^{(2(1-x))^{1+\frac{k}{n}}}\leq (1-x)^{(2x)}+x^{(2(1-x))}$$
So it seems here that we have a contradiction with :
$$Q(k+1)\Longrightarrow P(k-1)$$
Maybe we can exploit this to find a nice proof .
My questions :
It is good (I have a doubt at the end)?
Are the irrational exponent covered ?
Thanks in advance
Max
[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