Prove inequality $\ln \left( \frac{e-e^x}{1-x} \right) \leq \sqrt{\frac{1+x+x^2}{3}}$ for $0<x \le 1$

Question

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:

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.

Answer 1

$(A)\enspace$ 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.

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$(B)\enspace$ 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.

Answer 2

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 $h'''(0) \gt 0$.

Answer 3

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:

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

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