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Question: Are the properties as follows holds?

Version 1: the answer by Bjørn Kjos-Hanssen

Let $P$ be a positive integers. We written: $P=$ $a_1^{x_1}a_2^{x_2}...a_n^{x_n}$ $=b_1^{y_1}b_2^{y_2}...b_k^{y_k}$ where $a_i, b_j$ are integers greater than $1$.

if $\prod_{i=1}^n a_i \le \prod_{j=1}^k b_j $ then

1) $$\prod_{i=1}^n (1-\frac{1}{a_i}) \le \prod_{j=1}^k (1-\frac{1}{b_j}) $$

2) $$\prod_{i=1}^n (1-\frac{1}{a_i})^{x_i} \le \prod_{j=1}^k (1-\frac{1}{b_j})^{y_j} $$

3) $$\frac{\varphi(P)}{P} \le \prod_{i=1}^n (1-\frac{1}{a_i})$$

Where $\varphi(P)$ is the Euler's totient function of $P$

Comment: The answer of Bjørn Kjos-Hanssen is:

No, let $b_1,b_2,a_1,a_2$ be $2,6,3,4$. Then $$ (1-1/2)(1-1/6)=5/12 \not\ge 1/2=(1-1/3)(1-1/4) $$

But I think the asnswer is not counter example. Because $2.6 \le 3.4$ we can write

$$ (1-1/2)(1-1/6)=5/12 \le 1/2=(1-1/3)(1-1/4) $$.

To clearly I changes the question as follows:

Version 2:

Let $P$ be a positive integers. We written: $P=$ $a_1^{x_1}a_2^{x_2}...a_n^{x_n}$ $=b_1^{y_1}b_2^{y_2}...b_k^{y_k}$ where $a_i, b_j$ are integers greater than $1$.

if $\prod_{i=1}^n a_i < \prod_{j=1}^k b_j $ and $a_i \le b_j$ then

1) $$\prod_{i=1}^n (1-\frac{1}{a_i}) \le \prod_{j=1}^k (1-\frac{1}{b_j}) $$

2) $$\prod_{i=1}^n (1-\frac{1}{a_i})^{x_i} \le \prod_{j=1}^k (1-\frac{1}{b_j})^{y_j} $$

3) $$\frac{\varphi(P)}{P} \le \prod_{i=1}^n (1-\frac{1}{a_i})$$

Where $\varphi(P)$ is the Euler's totient function of $P$

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  • $\begingroup$ Can you kindly inform us about possible applications of these inequalities? Aren't the ones art for art's sake? TIA. $\endgroup$
    – user64494
    Commented Oct 10, 2019 at 10:16

2 Answers 2

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$2^46^118^1=3^34^116^1(=2^63^3)$.

$2\cdot6\cdot18=216>192=3\cdot4\cdot16$.

$(1/2)(5/6)(17/18)=85/216<15/32=(2/3)(3/4)(15/16)$.

Concerning the question about the phi-function, for $$n=2^{28}3^{16}=4\times6\times8\times9\times12\times16\times18\times24\times27\times32\times36\times48\times54$$ we have $${\phi(n)\over n}={1\over3}=.333\dots>.32956\dots={3\over4}\cdot{5\over6}\cdot\cdots\cdot{53\over54}$$

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  • $\begingroup$ I belive the Euler's totient function is minimum: en.wikipedia.org/wiki/Euler%27s_totient_function $\endgroup$
    – Đào Thanh Oai
    Commented Oct 10, 2019 at 9:39
  • $\begingroup$ I don't know what you mean. $\endgroup$
    – Gerry Myerson
    Commented Oct 10, 2019 at 9:39
  • $\begingroup$ I mean $\frac{\varphi(A)}{A} \le \prod_{i=1}^n (1-\frac{1}{a_i})$ Where $\varphi(A)$ is the Euler's totient function $\endgroup$
    – Đào Thanh Oai
    Commented Oct 10, 2019 at 9:46
  • $\begingroup$ But what is $A$? $\endgroup$
    – Gerry Myerson
    Commented Oct 10, 2019 at 9:48
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    $\begingroup$ I am sorry I mean $\frac{\varphi(P)}{P} \le \prod_{i=1}^n (1-\frac{1}{a_i})$ $\endgroup$
    – Đào Thanh Oai
    Commented Oct 10, 2019 at 9:59
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Answer to Version 1 of the question:

No, let $b_1,b_2,a_1,a_2$ be $2,6,3,4$. Then $$ (1-1/2)(1-1/6)=5/12 \not\ge 1/2=(1-1/3)(1-1/4). $$

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    $\begingroup$ In other words, the simplest possible example satisfying the hypothesis already fails to satisfy the conclusion. $\endgroup$
    – Gerry Myerson
    Commented Oct 10, 2019 at 5:39
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    $\begingroup$ do I want correct $\prod_{i=1}^n a_i \le \prod_{j=1}^k b_j $ to $\prod_{i=1}^n a_i < \prod_{j=1}^k b_j $ and $a_i \le b_j$. @GerryMyerson $\endgroup$
    – Đào Thanh Oai
    Commented Oct 10, 2019 at 6:39
  • $\begingroup$ Dear @BjørnKjos-Hanssen , because $2.6 \ge \le = 3.4$ so $(1-1/2)(1-1/6)=5/12 \not\ge 1/2=(1-1/3)(1-1/4)$ is not counter example. $\endgroup$
    – Đào Thanh Oai
    Commented Oct 10, 2019 at 6:55
  • $\begingroup$ @BjørnKjos-Hanssen I change my question. If You delete the answer, I will delete version 1 thank you. I am sorry. $\endgroup$
    – Đào Thanh Oai
    Commented Oct 10, 2019 at 7:29
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    $\begingroup$ It seems to me that if the conjecture with less-than-or-equal fell to the first possible counterexample, then it's a good idea to check for simple counterexamples before making a conjecture with less-than. There is no evidence that this has been done. Why present a conjecture with no reason to think it's true? $\endgroup$
    – Gerry Myerson
    Commented Oct 10, 2019 at 9:00

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