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The first function could be called 'exponential mean' of $y$ and $x$:

$$f(y,x)=\ln \left( \frac{e^y-e^x}{y-x} \right)$$

We can obtain it by Cauchy mean value theorem.

What is interesting, it appears numerically that:

$$\ln \left( \frac{e-e^x}{1-x} \right) \leq \sqrt{\frac{1+x+x^2}{3}}, \qquad 0<x \leq 1$$

$$\ln \left( \frac{e-e^x}{1-x} \right) \geq \sqrt{\frac{1+x+x^2}{3}}, \qquad x \geq 1$$

How would you prove both of these inequalities?

Around $1$ both functions are very close. See the plot below:

enter image description here

This is related to my recent question.

The series expansion doesn't seem to be the best way.

The first function can be represented:

$$\ln \left( \frac{e-e^x}{1-x} \right)=1+\ln(1-e^{-(1-x)})-\ln(1-x)$$

But that's not any better.

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  • $\begingroup$ I have inserted the proof for $0\leq x\le 1$. $\endgroup$ – user90369 Aug 30 '16 at 7:48
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$(A)\enspace$ Proof for $\,0\leq x\leq 1\,$:

Be $\,\displaystyle\sqrt{\frac{1+x+x^2}{3}}:=u^2+v^2\,$ and $\,\displaystyle\frac{1+x}{2}:=u^2-v^2\,$

with $\,\displaystyle v\in [0;\sqrt{\frac{1/\sqrt{3}-1/2}{2}}]\,$ and $\,\displaystyle u\in [\sqrt{\frac{1/\sqrt{3}+1/2}{2}};1]\,$.

Therefore is $\,\displaystyle\frac{1-x}{2}=\sqrt{3}2uv\,$.

The inequality can be written now: $$\frac{\sinh(2\sqrt{3}uv)}{ 2\sqrt{3}uv }\leq e^{2 v^2}$$

Because of $\,\displaystyle\frac{\sinh(x)}{x}\leq \frac{\sinh(y)}{y}\,$ for $\,0\leq x\leq y\,$ and $\,\max(u)=1$

it’s enough to show $\,\displaystyle\frac{\sinh(2\sqrt{3}v)}{ 2\sqrt{3}v }\leq e^{2 v^2}\,$.

With the series for $\,\sinh(x)\,$ and $\,e^x\,$ we get $\,\displaystyle\sum\limits_{k=0}^\infty \frac{(2\sqrt{3})^{2k}v^{2k}}{(2k+1)!}\leq \sum\limits_{k=0}^\infty \frac{2^k v^{2k}}{k!}\,$.

With $\,v^{2k}\geq 0\,$ and $\,\displaystyle\frac{6^k}{(2k+1)!}\leq \frac{1}{k!}\,$ for all $\,k\in\mathbb{N}_0\,$ follows the inequality.


$(B)\enspace$ Proof for $\,x\geq 1\,$ (using $\,u\,$ and $\,v\,$ as before but with different value ranges) :

We have for $\,x=1\,$ the values $\,v=0\,$ and $\,u=1\,$, for $\,x>1\,$ the values $\,v<0\,$ and $\,u>1\,$ .

The relation between $\,u\,$ and $\,v\,$ is given by $\,u^2-v^2+2uv\sqrt{3}=1\,$ and therefore

$u=\sqrt{1+4v^2}-v\sqrt{3}\geq 1\,$ for $\,v\leq 0\,$.

It’s more convenient to write $\,u=\sqrt{1+4z^2}+z\sqrt{3}\,$ with $\,z:=-v\geq 0\,$ .

Now we have to proof $\enspace\enspace\displaystyle e^{2z^2} \leq \frac{\sinh(2\sqrt{3}z( \sqrt{1+4z^2}+z\sqrt{3} ))}{2\sqrt{3}z(\sqrt{1+4z^2}+z\sqrt{3})} \enspace$ for $\,z\geq 0\,$ .

Due to $\enspace 1+z \leq \sqrt{1+4z^2}+z\sqrt{3}\enspace$ for $\,z\geq 0\enspace$ and $\enspace\displaystyle \frac{\sinh(x)}{x}\leq \frac{\sinh(y)}{y}\enspace$ for $\,0\leq x\leq y$

we can simplify the inequality to: $$e^{2z^2} \leq \frac{\sinh(2\sqrt{3}z(1+z))}{2\sqrt{3}z(1+z)}$$

It's $\enspace\displaystyle \frac{\sinh(2\sqrt{3}z(1+z))}{2\sqrt{3}z(1+z)} - e^{2z^2} =: \sum\limits_{k=0}^\infty z^{2k}(a_k+b_k) + \sum\limits_{k=0}^\infty z^{2k+1}c_k \geq 0\enspace$ because of:

$\displaystyle a_k := \frac{(2\sqrt{3})^{2k-2\large{\lfloor k/2\rfloor}}}{(2k-2\large{\lfloor k/2\rfloor}+1)!}{\binom {2k-2\large{\lfloor k/2\rfloor}}{2\large{\lfloor k/2\rfloor}}} - \frac{2^k}{k!}$

$\hspace{1.2cm} k:=2m$ : $\hspace{1.3cm}\displaystyle a_{2m}=\frac{(2\sqrt{3})^{2m}}{(2m+1)!} - \frac{2^{2m}}{(2m)!} \geq 0 $

$\hspace{4.3cm}$ <=> $\enspace 3^m \geq 2m+1\enspace$ , which is correct

$\hspace{1.2cm} k:=2m+1$ : $\hspace{0.5cm}\displaystyle a_{2m+1}=\frac{(2\sqrt{3})^{2m+2}}{(2m+3)!}{\binom {2m+2}{2m}} - \frac{2^{2m+1}}{(2m+1)!} \geq 0 $

$\hspace{4.3cm}$ <=> $\enspace 3^{m+1}(2m+1) \geq 2m+3\enspace$ , which is correct

$\displaystyle b_k = \sum\limits_{n=0}^{\large{\lfloor k/2\rfloor}-1} \frac{(2\sqrt{3})^{2k-2n}}{(2k-2n+1)!}{\binom {2k-2n}{2n}} \geq 0 \enspace$ , $\enspace b_k>0\,$ for $\,k\geq 2\,$

$\displaystyle c_k = \sum\limits_{n=0}^{\large{\lfloor (k-1)/2\rfloor}} \frac{(2\sqrt{3})^{2k-2n}}{(2k-2n+1)!}{\binom {2k-2n}{2n+1}} \geq 0 \enspace$ , $\enspace c_k>0\,$ for $\,k\geq 1\,$

$a_0=b_0=b_1=c_0=0$

This concludes the proof.

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  • $\begingroup$ Thank you. I'll check this in delail at a later time, but looks good so far $\endgroup$ – Yuriy S Aug 30 '16 at 7:51
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So my solution is a bit of "brute force" solution, not very elegant and nice. But that was how I was tackling algebra problems when I can't find better solution. First let's assume: $$ l(x) = \ln\left(\frac{e - e^x}{1-x}\right) $$ $$ l(1) = \lim_{x \to 1} l(x) = \ln \lim_{x \to 1} \frac{e - e^x}{1-x} $$ $$ l(1) = \ln \lim_{x \to 1} \frac{-e^x}{-1} = \ln(e) = 1$$ $$ f(x) = 3l(x)^2 $$ $$ g(x) = x^2+x+1 $$ $$ h(x) = f(x) - g(x) $$

Idea:

Show that $h'(x) \ge 0$ for $x \gt 0$ and that $h(1)=0$. This implies that $h(x)$ is monotonically increasing. Using the fact that $h(1)=0$ it implies that $f(x) \lt g(x)$ for $x \lt 1$ and $f(x) \gt g(x)$ for $x \gt 1$.

Note that $l(x) \gt 0$ and $x^2+x+1 > 0$ for $x \gt 0$, which means that the inequality for $f$ and $g$ implies the inequality for the original functions (e.g. we just square the initial functions).

Proof: $$l'(x) = \frac{1}{1-x} - \frac{1}{e^{1-x}-1}$$ $$l'(1) = \lim_{x \to 1} \frac{e^{1-x} - (1-x)}{(1-x)(e^{1-x}-1)} = \lim_{x \to 1} \frac{1 -e^{1-x}}{(1-e^{1-x}) + (x-1)e^{1-x}}$$ $$l'(1) = \lim_{x \to 1} \frac{e^{1-x}}{e^{1-x} + e^{1-x} + (1-x)e^{1-x}} = \lim_{x \to 1} \frac{1}{3 - x} = \frac{1}{2}$$ $$f'(x) = 6 l(x) l'(x)$$ $$g'(x) = 2x+1 $$ $$h'(x) = f'(x) - g'(x)$$

We want to show that $h'(x) \ge 0$. A sufficient condition for this is that $h''(x) = 0$ has a unique solution $x^*$ for $ x \gt 0$ and that it is a monotonically increasing function ($h''(x)$). The condition implies that $h'(x)$ has a unique minimum and that is $x^*$. Furthermore, we will show that $h'(x^*) = 0$.

$$l''(x) = \frac{1}{(1-x)^2} - \frac{e^{1-x}}{(e^{1-x}-1)^2}$$ $$l''(1) = \lim_{x \to 1} \frac{(e^{1-x}-1)^2 - (1-x)^2 e^{1-x}}{(1-x)^2e^{1-x}-1)^2} = \frac{1}{12}$$ $$f''(x) = 6l(x)l''(x) + 6 l'(x)^2$$ $$g''(x) = 2$$ $$h''(x) = 6l(x)l''(x) + 6 l'(x)^2 - 2 $$ $$ h''(x) = 0 \iff l(x)l''(x) + l'(x)^2 = \frac{1}{3} $$

Now to proof this first note that $h''(1) = \frac{1}{3}$, thus our candidate $x^*=1$. Also, as expected $h'(1)=0$. To conclude the proof we need to show that $h''(x)$ is monotonically increasing for $x \gt 0$.

For now I need to prove it, but if you plot the function it looks like it is. My guess is you will need to proof in fact that $h'''(0) \gt 0$.

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  • $\begingroup$ Thank you for the nice work! $\endgroup$ – Yuriy S Aug 27 '16 at 15:27
  • $\begingroup$ So far it doesn't seem that the last part will be easy to proof. There are two approaches to proove it. First, find some inequalities about logarithms and maybe things will solve out. If not, another approach, which I think you mentioned, is to use Taylor expansion and show that h''(x+eps) > h''(x) for any eps. I will attempt one of these at some point and see what comes out. $\endgroup$ – Alex Botev Aug 27 '16 at 15:34
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A little demonstration (not a full solution, but I'll try to expand it later):

If we were allowed to use CAS, such as Mathematica (or spent a little time differentiating), we could just square both sides and expand the left side into series aroud $x=1$:

$$\ln^2 \left( \frac{e-e^x}{1-x} \right) =1+(x-1)+\frac{1}{3} (x-1)^2+\frac{1}{24} (x-1)^3+\frac{1}{960} (x-1)^4+O((x-1)^5)$$

Now we expand and simplify the first three terms:

$$\ln^2 \left( \frac{e-e^x}{1-x} \right) =\frac{1}{3}+\frac{x}{3}+\frac{x^2}{3}\color{blue}{+\frac{1}{24} (x-1)^3+\frac{1}{960} (x-1)^4}+O((x-1)^5)$$

Comparing with the square of the right-hand side we immediately see:

$$\ln^2 \left( \frac{e-e^x}{1-x} \right)- \frac{1+x+x^2}{3}=\frac{1}{24} (x-1)^3+\frac{1}{960} (x-1)^4+O((x-1)^5)$$

Thus, around $x=1$ we have:

$$\ln^2 \left( \frac{e-e^x}{1-x} \right)- \frac{1+x+x^2}{3}=-\frac{1}{24} (1-x)^3+\frac{1}{960} (1-x)^4-O((1-x)^5) \leq 0, $$ $$x \to 1^-$$

$$\ln^2 \left( \frac{e-e^x}{1-x} \right)- \frac{1+x+x^2}{3}=\frac{1}{24} (x-1)^3+\frac{1}{960} (x-1)^4+O((x-1)^5) \geq 0, $$ $$ x \to 1^+$$

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