I study about primorial.
I found some property in wiki; [Applications and properties] in https://en.wikipedia.org/wiki/Primorial. That is "Here For each primorial n, the fraction $\frac{\phi(n)}{n}$ is smaller than it for any lesser integer, where $\phi$ is the Euler totient function."
Actually I want to obtain the proposition: For $m<p_{n}\sharp$, then $\frac{\phi(p_{n}\sharp)}{p_{n}\sharp}<\frac{\phi(m)}{m}$ where positive integer $n$, $m$ and $\phi$ is Euler's function .
The problem is:
(1) Are wiki's argue and the proposition what I want same?
(2-1) If same, Where from the wiki's argue?
(2-2) If not same, is there any clue for the proposition?
First, note that if $p_i\mid m$ for all $1\le i \le n$, then $m\ge \prod_{i=1}^n(p_i)=p_n\#$. So if we choose $m<p_n\#$, then there must be one or more values of $i$ such that $p_i\not\mid m$.
Let us divide the set of the first $n$ prime numbers $\{p_n\}$ into two subsets: $\{p_j\}$ such that $p_j\not\mid m$ and $\{p_k\}$ such that $p_k\mid m$. Now we can say $$p_n\#=\prod(p_j)\prod(p_k)\\ \frac{\phi(p_n\#)}{p_n\#}=\prod\frac{p_j-1}{p_j}\prod\frac{p_k-1}{p_k}$$ Also, $$m=\prod (p_k)^{a_k} \\ \frac{\phi(m)}{m}=\prod\frac{p_k^{a_k-1}(p_k-1)}{(p_k)^{a_k}}=\prod\frac{p_k-1}{p_k}$$ Substituting the last result in the first, we see $$\frac{\phi(p_n\#)}{p_n\#}=\prod\frac{p_j-1}{p_j}\frac{\phi(m)}{m}$$ We know there is at least one term in the product over the index $j$, and that for every such term $\frac{p_j-1}{p_j}<1$
Hence we can say that for $m<p_n\#$, $$\frac{\phi(p_n\#)}{p_n\#}<\frac{\phi(m)}{m}$$