Stronger statement for $x^{2(1-x)}+(1-x)^{2x}\leq 1$

Question

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

Answer 1

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 $$\left( \begin{array}{cccc} n & x_{\text{min}}& f_{\text{min}} & \text{method}\\ 2 & 0.213073 & 0.990668 & \text{Newton} \\ 3 & 0.217393 & 0.990665 & \text{Halley} \\ 4 & 0.216372 & 0.990665 & \text{Householder} \\ 5 & 0.216472 & 0.990665 & \text{no name} \\ 6 & 0.216452 & 0.990665 & \text{no name} \\ 7 & 0.216454 & 0.990665 & \text{no name} \end{array} \right)$$

Answer 2

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.