Problem :
Let $0<x<1$ and $k\geq 1$ then we have :
$$x^{(2(1-x))^{k}}+(1-x)^{(2x)^{k}}\leq 1$$
My strategy or sketch of a proof :
Part one the case $0<x\leq 0.25$
Using Bernoulli's inequality we have :
$$(1-x)^{(2x)^{k}}\leq 1-x(2x)^k$$
So we need to show :
$$x^{(2(1-x))^{k}}\leq x(2x)^k$$
We can finish it using derivatives .
Part two the case $k\geq 1+\alpha$ :
We define $\alpha$ as :
$$\lim_{x\to 0.5^-}=\frac{\ln\Big(2(2^{2x}x^2(1-x)2)\Big)}{\ln(2x)}=\frac{\ln\Big(2(1-(2^{2x}x^2(1-x)2))\Big)}{\ln(2(1-x))}=\alpha=1.69315\cdots$$
We have :
$$x^{2(1-x)}\leq2^{2x}(1-x)2x^2$$
And :
$$(1-x)^{2x}\leq 1-2^{2x}(1-x)2x^2$$
For a sketch of proof see Refinements of the inequality $f(x)=x^{2(1-x)}+(1-x)^{2x}\leq 1$ for $0<x<0.5$
So we have :
$$x^{(2(1-x))^k}\leq f(x)^{(2(1-x))^{k-1}}=(2^{2x}(1-x)2x^2)^{(2(1-x))^{k-1}}$$
And :
$$(1-x)^{(2x)^k}\leq (1-2^{2x}(1-x)2x^2)^{(2x)^{k-1}}$$
Now we want $y\geq 1$ a real number :
$$(2x)^{k-1}\geq (2f(x))^{y}$$ And $$(2(1-x))^{k-1}\geq (2(1-f(x)))^{y}$$
Solving this we have
$$\frac{\ln\Big(2(1-(2^{2x}x^2(1-x)2))\Big)}{\ln(2(1-x))}\leq \frac{k-1}{y}\leq \frac{\ln\Big(2(2^{2x}x^2(1-x)2)\Big)}{\ln(2x)}\quad (1)$$
The constraint $(1)$ is honoured if :$$\alpha=\frac{k-1}{y}$$
So we have :
$$x^{(2(1-x))^k}\leq (f(x))^{(2(1-f(x))^{\frac{k-1}{\alpha}} }\quad (2)$$
And :
$$(1-x)^{(2x)^k}\leq (1-f(x))^{(2f(x))^{\frac{k-1}{\alpha}}}\quad (3)$$
Adding $(2)$ and $(3)$ we have :
$$h(k)\leq (1-f(x))^{(2f(x))^{\frac{k-1}{\alpha}}}+(f(x))^{(2(1-f(x))^{\frac{k-1}{\alpha}}}$$
We can repeat the reasoning $p$ times ($p\geq 1$ a natural number) and the last exponent becomes :
$$1\leq \frac{k-1}{\alpha^p}\leq \alpha$$
It's true because we can always choose $k$ such that :
$$\alpha^p\leq k \leq \alpha^{p+1}$$
So we reduce this case to the case $1\leq k\leq 1+\alpha $
Part 3 the case $k\leq 1+\alpha $ and $0.25<x\leq 0.5$
Recently user RiverLi showed the lemma :
Lemma
Let $1< x$ and $a>2$ then we have :
$$r(x)=\left(1+\frac{\left(\ln\left(ax\right)^{c}-\ln\left(a\right)^{c}\right)}{\left(\ln\left(2a\right)^{c}-\ln\left(a\right)^{c}\right)}-x\right)\leq 0$$
We can find $c$ as $r'(2)=0$.
We cannot apply it directly but it's possible with the new form :
$$\left(x^{-(2(1-x))^{k}}-1\right)\left((1-x)^{-(2x)^{k}}-1\right)\geq 1$$
Now we apply twice the lemma and we start by introducing (see the reference) :
$$f(k)=((\ln((a)x^{-(2(1-x))^k}))^c-(\ln(a))^c) ((\ln((a)(1-x)^{-(2(x))^k}))^c-(\ln(a))^c)$$
Conjecture :
Let $a=10^{10}$ in $f(k)$ then the function $g(k),h(k)$ are log convex always with the contraint above where :
$$g(k)=((\ln((a)x^{-(2(1-x))^k}))^c-(\ln(a))^c) $$
And
$$l(k)=((\ln((a)(1-x)^{-(2(x))^k}))^c-(\ln(a))^c)$$
So it reduces to the case $k=1$ wich is the hardest here with the new form .
Two questions :
1)Is it good I mean nothing wrong here ?
2)How to show the missing parts especially the case $k=1$ where I used the lemma ?
Thanks you!
Reference:
[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
https://link.springer.com/article/10.1186/1029-242X-2013-468
