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This is an inequality based on an idea from Michael Rozenberg in this question. This is the following:

Let $x,y,z$ be positive real numbers with $xyz=1$ then we have:

$$\frac{0.5\sqrt{\pi}}{\Gamma(f(x))}+\frac{0.5\sqrt{\pi}}{\Gamma(f(z))}+\frac{0.5\sqrt{\pi}}{\Gamma(f(y))}\leq 1.5$$

With $f(x)=\dfrac{x}{x^{11}+1}$.

Sincerely, I have no idea to prove this because this is more difficult than the original one.

Any help is appreciated!

Thanks a lot.

Botond
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1 Answers1

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FINAL EDITION $08.03.18$

FUNCTIONS

At first, let us consider the function $$f(t)=\dfrac{t}{t^{11} + 1},\quad t\in[0, \infty).$$ Easy to see that $$f(0) = f(\infty) = 0$$ and $$\dfrac{df}{dt} = \dfrac{1 - 10 t^{11}}{(t^{11} + 1)^2}.$$ So $f(t)$ is unimodal in the interval $[0, \infty),$ with the maximum $$f'(t_m) = 0,\quad t_m = \dfrac1{\sqrt[11]{10}} \approx 0.81113,\quad f_m = f(t_m) = \dfrac1{11}10^{10/11}\approx 0.73739.\tag1$$ This means that the actual range of values of the gamma function is $(0,f_m),$ where the terms increases monotonically and $$g_m = \dfrac{\sqrt\pi}{\Gamma(f_m)} \approx 1.4264505975\dots\tag2$$

Also note that $$f(1) = \dfrac 12,\quad \Gamma\left(\dfrac12\right) = \sqrt\pi.$$

This means that the issue inequality becomes exact equality when $$x = y = z = 1,\quad f(x)= \dfrac12.$$ In the same time, if $t>1,$ then $f(t) < \dfrac12.$

Thus, at least one of values $x,\ y,\ z$ belongs to the interval $(0, 1].$

These are the reasons to consider the function $$v(t) = \dfrac {\sqrt\pi}{\Gamma\left(t+\dfrac12\right)}\tag3$$ for the values $t+\dfrac12 \in [0,\ f_m],$ where it increases monotonically.

Then, one can obtain the Taylor series in the form of $$v(t) = 1 + v_1t + v_2t^2 + v_3t^3 + v_4t^2 +\dots,\tag4$$ where \begin{aligned} &v_1 = -\psi^{(0)}\left(\dfrac12\right) = \gamma + \log 4 \approx1.9635100260\dots,\\ &v_2 = -\dfrac14(\pi^2 - 2v_1^2) = -\dfrac14(p - 2\pi^2) \approx 0.53971 52891\dots,\\ &v_3 = -\dfrac1{12}(3\pi^2v_1 - 2v_1^3 - 28\zeta(3)) = -\dfrac1{12}(pv_1-28\zeta(3))\approx -0.7782905521\dots,\\[4pt] &v_4 = \dfrac1{96}(4v_1^4 - 12\pi^2v_1^2 + 224\zeta(3)v_1)\\[4pt] &\quad = \dfrac1{96}(p^2 - 10\pi^2 - 8\cdot28\zeta(3)v_1)\approx 0.3555284642\dots,\\ &\gamma \approx 0.5772156649\dots \text{(Euler gamma constant)},\\[4pt] &p = 3\pi^2 - 2v_1^2 \approx 21.8980699586\dots,\\[4pt] &28\zeta(3) = 33.6575932884\dots. \end{aligned} Easy to see that $$v(t) \le 1 + v_1\left(t-\dfrac12\right)\tag5$$

Taking in account the proof of Michael Rozenberg inequality $$f(x) + f(y) + f(z) \le \dfrac32\quad \text{ if } (x,y,z)\in(0,\infty)^3 \wedge xyz = 1,$$

one can write: $$v(f(x)) + v(f(y)) + v(f(z)) - 3 \le v_1\left(f(x) -\dfrac12 + f(y) -\dfrac12 + f(z) -\dfrac12\right) \le 0,$$

$$\boxed{v(f(x)) + v(f(y)) + v(f(z)) \le 3.} $$

Yuri Negometyanov
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  • I think it's very interesting because you have separated the problem in another wich is simpler . Ps : at the end it's not $xyz=1$ ? – max8128 Feb 28 '18 at 15:06
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    Can you elaborate a bit on the step (4) -> (5)? Unless I made an error, $h(1) \approx 0.94$, so that for $x=y=z=1$ $h(x)+h(y)+h(z) \approx 2.82$, which is larger than $3/\sqrt \pi \approx 1.69$. – Martin R Feb 28 '18 at 15:15
  • @MartinR yes I remark that to .In fact without any numerical argument there is a strict inequality follow by a larger one . So there is a problem... – max8128 Feb 28 '18 at 15:20
  • Indeed, the original inequality is sharp, so any solution with strict inequalities cannot be correct. – Martin R Feb 28 '18 at 15:37
  • @MartinR This is only idea, not the final edition. The approximation will be changed. Thanks, you are right. – Yuri Negometyanov Feb 28 '18 at 15:46
  • There is so many problems that...First you can invoke a more general concept than the unimodality like the quasiconvexity and use a Jensen like formula . Secondly you can't cut the Taylor series it involves the last message of MartinR . Finally I think you can't solve this problem without separate the problem in different cases . However good luck ! –  Feb 28 '18 at 16:02
  • @FatsWallers I can fix all the problems, using more hard way. I'm doing that now, but I would not update the answer permanently. – Yuri Negometyanov Feb 28 '18 at 23:15
  • I trust in you man thanks for your interest ! –  Mar 01 '18 at 11:49
  • @FatsWallers The question is interesting. And this is a challenge for me. – Yuri Negometyanov Mar 01 '18 at 13:37
  • @MartinR And now? – Yuri Negometyanov Mar 02 '18 at 08:54