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Post Closed as "Needs details or clarity" by Mark Wildon, R.P., Ilya Bogdanov, LeechLattice, Tom De Medts
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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. $$\frac{\varphi(A)}{A} \le \prod_{i=1}^n (1-\frac{1}{a_i})$$$$\frac{\varphi(P)}{P} \le \prod_{i=1}^n (1-\frac{1}{a_i})$$

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

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

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

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

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

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

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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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. $$\frac{\varphi(A)}{A} \le \prod_{i=1}^n (1-\frac{1}{a_i})^{x_i}$$$$\frac{\varphi(A)}{A} \le \prod_{i=1}^n (1-\frac{1}{a_i})$$

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

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

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

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. $$\frac{\varphi(A)}{A} \le \prod_{i=1}^n (1-\frac{1}{a_i})^{x_i}$$

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

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

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

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

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

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

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

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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

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

  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. $$\frac{\varphi(A)}{A} \le \prod_{i=1}^n (1-\frac{1}{a_i})^{x_i}$$

$$\prod_{i=1}^n (1-\frac{1}{a_i})^{x_i} \le \prod_{j=1}^k (1-\frac{1}{b_j})^{y_j} $$ Where $\varphi(A)$ is the Euler's totient function

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

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

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

$$\prod_{i=1}^n (1-\frac{1}{a_i})^{x_i} \le \prod_{j=1}^k (1-\frac{1}{b_j})^{y_j} $$ Where $\varphi(A)$ is the Euler's totient function of $A$

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

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

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

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

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

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

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. $$\frac{\varphi(A)}{A} \le \prod_{i=1}^n (1-\frac{1}{a_i})^{x_i}$$

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

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

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

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