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Hi it's related to Showing the inequality $f(x)=x^{2(1-x)}+(1-x)^{2x}\leq 1$ for $0<x<1$

We want to show 1:

Let $0<x<0.5$ 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 :

$$p(x)+h(x)\leq 1$$

Wich is equivalent to :

$$-2(x-1)x((4^x-2)x+(2x-1)\ln(1-x))\leq 0$$

It's not hard so I omitt here the proof of this fact .

We are done .

Question :

Is it right ?

Thanks in advance .

Regards 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

Alex Ravsky
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Miss and Mister cassoulet char
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1 Answers1

3

What you've done looks correct to me.

$$g(x)=\ln\Big(x^{2(1-x)}\Big)-\ln\Big(2^{2x+1}(1-x)x^2\Big)$$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.

I think that this is correct. (A proof that $g'(x)\gt 0$ : We have $g'(x)=\dfrac{ F(x)}{1-x}$ where $F(x)=x (2 + \ln(4)) + 2 (x - 1) \ln (x) - 1 - \ln(4)$, $F'(x)=\underbrace{\dfrac{2(2x-1)}{x}}_{\lt 0}+\underbrace{\ln(4x^2)}_{\lt 0}\lt 0$ and $F(0.5)=0$. These imply $F(x)\gt 0$ from which $g'(x)\gt 0$ follows.)

For the Rhs I use one of the lemma (7.1) due to Vasile Cirtoaje we have :$$(1-x)^{2x}\leq p(x)=1-4(1-x)x^{2}-2(1-x)x(1-2 x)\ln(1-x)$$

I think that you have correctly got the inequality by setting $a=1-x$ and $c=2x$ in the lemma.

So we have :$$q(x)\leq p(x)+h(x)$$We want to show that :$$p(x)+h(x)\leq 1$$Wich is equivalent to :$$-2(x-1)x((4^x-2)x+(2x-1)\ln(1-x))\leq 0$$

I've found no errors here. (I don't check the last inequality since you've written "It's not hard so I omitt here the proof of this fact".)

mathlove
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