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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

Barackouda
  • 3,879

1 Answers1

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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.

River Li
  • 49,125