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For all $a,b>0$ define $$ G=\sqrt{ab},\quad A=\frac{a+b}{2},\quad L=\begin{cases}a&a=b\\\frac{b-a}{\log b-\log a}&a\ne b\end{cases}. $$ In the "Logarithmic Mean Revisited" paper, it is shown that $$ L \leq \frac{2}{3}G+\frac{1}{3}A. $$ From https://kuing.cjhb.site/forum.php?mod=redirect&goto=findpost&pid=56797&ptid=11685 : By Power Mean Inequality (Jensen's inequality applied to $f(x)=x^{4/5}$) we have $(\frac{2}{3}G^{4/5}+\frac{1}{3}A^{4/5})^{5/4}≤\frac{2}{3}G+\frac{1}{3}A$ so an improvement of the inequality would be: $$\tag1\label{1} L^{4/5}\leq\frac{2}{3}G^{4/5}+\frac{1}{3}A^{4/5}, $$ Questions:

  • Is this inequality true?
  • Is the exponent $4/5$ optimal (i.e. it cannot be replaced by any smaller number)?

Some ideas: Consider a change of variables by setting $x = \frac{b}{a} > 0$ to reduce \eqref{1} to a single-variable form $$ \left(\frac{x-1}{\log x}\right)^{4/5}\leq \frac{2}{3}x^{2/5} + \frac{1}{3}\left(\frac{x+1}{2}\right)^{4/5}. $$ The substitution $x = e^{2y}$ transforms this into $$ \left(\frac{e^{2y}-1}{2y}\right)^{4/5}\leq \frac{2}{3}e^{4y/5} + \frac{1}{3}\left(\frac{e^{2y}+1}{2}\right)^{4/5}. $$ Dividing both sides by $e^{4y/5}$, we get $$ \left(\frac{\sinh y}y\right)^{4/5}\leq \frac{2}{3} + \frac{1}{3}\left(\cosh y\right)^{4/5}. $$ Rearranging $$ \left(\left(\frac{\sinh y}y\right)^{4/5} - \frac{2}{3}\right)^5 \leq \frac{1}{3^5}\left(\cosh y\right)^4. $$ Both $\frac{\sinh y}{y}$ and $\cosh y$ are even functions. It suffices to prove the $n$th coefficient of Taylor series at $y=0$ of LHS is less than that of RHS for all even $n>0$.

Using $\frac{\sinh y}{y}=1+\frac{y^2}{3!}+\frac{y^4}{5!}+\cdots$, the $n$th coefficient of Taylor series at $y=0$ of LHS is $$??\label{2}\tag{2}$$ Using $(\cosh y)^4=\frac{1}{8} (4 \cosh (2 y)+\cosh (4 y)+3)$, the $n$th coefficient of Taylor series at $y=0$ of RHS is $$\frac{2^{n-3}(4 + 2^n)}{3^5n!}.\label{3}\tag{3}$$for all even $n>0$.

hbghlyj
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    When you reduce to the single-variable form, don't you need to raise the left side to $\frac45$? – J. W. Tanner Mar 21 '25 at 01:44
  • @J.W.Tanner Yes. Sorry, I made a silly mistake. – hbghlyj Mar 21 '25 at 01:45
  • You have to count only $x>1$(that is, $y>0$) because $A,G,L$ are all symmetric and $x=1$ case is so trivial. And your idea, comparing coefficient of Talyor series, is much more strong than comparing values. For example, $f(x)=x+x^2-x^3+x^4$ is greater than 0 for all $x>0$ but 3rd coefficient of Talyor series of $f$ is negative. – Myungheon Lee Mar 21 '25 at 02:03
  • Sorry, the left hand side $\left(\left(\frac{\sinh y}y\right)^{4/5} - \frac{2}{3}\right)^5 $ is not analytic at 0 since it contains fractional powers. – hbghlyj Mar 21 '25 at 02:10
  • @hbghlyj Your inequality is very strong. My congratulations! – Michael Rozenberg Mar 21 '25 at 06:26

2 Answers2

5

Sketch of a proof.

It suffices to prove that, for all $x \ge 1$, $$\frac{x-1}{\ln x}\leq \left(\frac{2}{3}x^{2/5} + \frac{1}{3}\left(\frac{x+1}{2}\right)^{4/5}\right)^{5/4}, $$ or $$\ln x \ge \frac{x - 1}{\left(\frac{2}{3}x^{2/5} + \frac{1}{3}\left(\frac{x+1}{2}\right)^{4/5}\right)^{5/4}}.$$

Letting $y := \sqrt[5]{x}$, it suffices to prove that, for all $y \ge 1$, $$5\ln y \ge \frac{y^5 - 1}{\left(\frac{2}{3}y^2 + \frac{1}{3}\left(\frac{y^5+1}{2}\right)^{4/5}\right)^{5/4}}. \tag{1}$$

We use the following bound: $$\frac{2}{3}y^2 + \frac{1}{3}\left(\frac{y^5+1}{2}\right)^{4/5} \ge \left(\frac{7y^3 + 5y^2 + 5y + 7}{12y^2 + 12}\right)^4, \quad \forall y \ge 1. \tag{2}$$ (Note: Take logarithm, then take derivative.)

To prove (1), using (2), it suffices to prove that, for all $y \ge 1$, $$5\ln y \ge \frac{y^5 - 1}{\left(\frac{7y^3 + 5y^2 + 5y + 7}{12y^2 + 12}\right)^5}. \tag{3}$$ (3) is true (taking derivative).

About $4/5$ to be optimal.

Let $r > 0$ be fixed. Let $$f(x) := \frac23 x^{r/2} + \frac13 \left(\frac{x + 1}{2}\right)^r - \left(\frac{x - 1}{\ln x}\right)^r.$$

We have $$\lim_{x\to 1^{+}} f' (x) = \lim_{x\to 1^{+}} f'' (x) = \lim_{x\to 1^{+}} f'''(x) = 0, $$ $$\lim_{x\to 1^{+}} f^{(4)} (x) = \frac{1}{120}r(5r - 4).$$

Thus, $$\lim_{x\to 1^{+}} \frac{f(x)}{(x-1)^4} = \frac{1}{2880}r(5r-4).$$

River Li
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It is probably better to use $$\sinh x/x=\prod_{n=1}^{\infty}(1+(x^2/n^2\pi^2))$$ and $$\cosh x=\prod_{n=1}^{\infty}(1+(4x^2/(2n-1))^2\pi^2)$$ and to try to prove for $p=4/5$ $$3<\frac{2x^p}{\sinh^p x}+\prod_{n=1}^{\infty}\left(1+\frac{(4n-1)x^2}{n^2(2n-1)^2(n^2\pi^2+x^2)}\right)^p.$$

hbghlyj
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Letac Gérard
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