Minimal value for the specific summatory Euler Phi function Let $H(q)$ be the set of reduced residues $mod(q)$ and $\Phi(a)$ Euler totient function. How can I evaluate
$\min_{q\leq x}\frac{1}{q}\sum_{a\epsilon\ H(q)}\frac{\Phi (a)}{a}$
 A: Let us find a good approximation for your sum for a given large $q$. I will use the notation $\varphi(a)$ for the Euler function. First of all, by Möbius inversion,
$$
\sum_{a\in H(q)}\frac{\varphi(a)}{a}=\sum_{a\leq q}\left(\sum_{d\mid (a,q)}\mu(d)\right)\frac{\varphi(a)}{a}=\sum_{d\mid q}\mu(d)S_d(q),
$$
where
$$
S_d(q)=\sum_{bd\leq q}\frac{\varphi(bd)}{bd}.
$$
To evaluate this sum, notice that
$$
\frac{\varphi(m)}{m}=\sum_{d\mid m}\frac{\mu(d)}{d},
$$
hence
$$
S_d(q)=\sum_{bd\leq q}\sum_{d_1\mid bd}\frac{\mu(d_1)}{d_1}=\sum_{d_1\leq q}\frac{\mu(d_1)}{d_1}\#\{bd\leq q: d_1\mid bd\}.
$$
Now, $d_1\mid bd$ iff $\frac{d_1}{(d,d_1)}\mid b$, so the weight inside our sum is
$$
\left[\frac{q/d}{d_1/(d,d_1)}\right]=\frac{q(d,d_1)}{dd_1}+O(1).
$$
Therefore,
$$
S_d(q)=\frac{q}{d}\sum_{d_1\leq q}\frac{\mu(d_1)(d,d_1)}{d_1^2}+O(\ln q)
$$
(the constant in O is absolute)
The sum in the main term is
$$
\sum_{d_1\leq q}\frac{\mu(d_1)(d,d_1)}{d_1^2}=\sum_{d_1}\frac{\mu(d_1)(d,d_1)}{d_1^2}+O\left(\sum_{d_1>q}\frac{d}{d_1^2}\right)=\prod_{p}\left(1-\frac{(p,d)}{p}\right)+O(dq^{-1}),
$$
because $(d,d_1)$ is multiplicative as a function of $d_1$. Thus,
$$
S_d(q)=\frac{q}{d}\prod_{p}\left(1-\frac{(p,d)}{p}\right)+O(\ln q)=\frac{q}{d}\prod_{p\mid d}\left(\frac{p}{p+1}\right)\prod_p\left(1-\frac{1}{p^2}\right)+O(\ln q)=
$$
$$
=\frac{6q}{\pi^2 d}\prod_{p\mid d}\frac{p}{p+1}+O(\ln q).
$$
Plugging this into the initial formula, we obtain
$$
\sum_{a\in H(q)}\frac{\varphi(a)}{a}=\frac{6q}{\pi^2}\sum_{d\mid q}\frac{\mu(d)}{d}\prod_p \frac{p}{p+1}+O(\sigma_0(q)\ln q).
$$
Using multiplicativity once more, we get
$$
\sum_{a\in H(q)}\frac{\varphi(a)}{a}=\frac{6q}{\pi^2}\prod_{p\mid q}\left(1-\frac{1}{p+1}\right)+O(\sigma_0(q)\ln q).
$$
So, up to an error of order $q^{-1+o(1)}$ the quantity you want to minimise is
$$
\frac{6}{\pi^2}\prod_{p\mid q}\left(1-\frac{1}{p+1}\right),
$$
therefore your guess about primorials above is essentially correct and the minimum is asymptotic to
$$
\frac{6}{\pi^2}\prod_{p\leq \ln x}\left(1-\frac{1}{p+1}\right)=\frac{6}{\pi^2}\prod_{p\leq \ln x}\left(1-\frac{1}{p}\right)\left(1-\frac{1}{p^2}\right)^{-1}\sim \frac{e^{-\gamma}}{\ln\ln x}.
$$
