We can start from the explicit expression of $\displaystyle \varphi (n)$ ...
$\displaystyle \varphi(n)= n\cdot \prod_{p|n} (1-\frac{1}{p})$ (1)
... from which we derive immediately...
$\displaystyle \frac{\varphi(n)}{n} = \prod_{p|n} (1-\frac{1}{p})$ (2)
Now we remember the well known 'Euler's product' written in the 'not very usual form' ...
$\displaystyle \frac{1}{\zeta(s)} = \prod_{p} (1-\frac{1}{p^{s}})$ (3)
... from which, setting $\displaystyle s=1$ we derive immediately...
$\displaystyle \prod_{p} (1-\frac{1}{p})=0$ (4)
Doesn't Riemann's zeta function has a pole in s = 1?
Tonio
The (4) tells us that for any $\displaystyle \varepsilon >0$ we have an integer $\displaystyle k$ so that is...
$\displaystyle \prod_{n=1}^{k} (1-\frac{1}{p_{n}}) < \varepsilon$ (5)
Combining (2) and (5) we conclude that for any $\displaystyle \varepsilon >0$ it exist a k so that setting...
$\displaystyle n=\prod_{i=1}^{k} p_{i}$ (6)
...is...
$\displaystyle \frac{\varphi(n)}{n} < \varepsilon$ (7)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$