Let $\phi_a(n)= 1$ if $\phi(n) \geq \frac{n}{a}$ and $0 $ otherwise. where $\phi(n)$ is Euler phi function.
Let $S_a(x) = \sum \limits_{n \leq x} \phi_a(n)$ and $M_a = \lim \limits_{x \to \infty}\frac{S_a(x)}{x} $ if the limit exists.
Trivially we have $M_a \leq 1$ for all $a \in \mathbb{R_{+}}$.
The lower bound trivially is $M_a \geq 0$, with a little work we have $ \sum \limits_{n \leq x} \frac{\phi(n)}{n} \approx \frac{6}{\pi^2}x$ so we can say that $\sum \limits_{n \leq x , \phi_a(n)=1} \frac{\phi(n)}{n} + \sum \limits_{n \leq x , \phi_a(n)=0} \frac{\phi(n)}{n} \leq \sum \limits_{n \leq x , \phi_a(n)=1} 1 + \sum \limits_{n \leq x , \phi_a(n)=0} \frac{1}{a} = S_a(x) +\frac{x-S_a(x)}{a} $
So we have that $S_a(x)+\frac{x-S_a(x)}{a} \geq \frac{6}{\pi^2}x $ and so $M_a (1-\frac{1}{a}) \geq \frac{6}{\pi^2}-\frac{1}{a} $ so for $a\geq 2$ we have that $M_a \geq \frac{\frac{6}{\pi^2}-\frac{1}{a}}{1-\frac{1}{a}}$
For instance $a=2$ gives that $M_2 \geq 0.11773 $
Its better than the trivial bound but still not quite good, is there a more cleaver way to find $M_a$ ?
For all $a> 1$ since for $0< a\leq 1$ we have that $M_a=0$