In their paper, On the distribution of reduced residues, Montgomery and Vaughan state early on that

*With a more careful argument from (2) it is easily seen that
$$\tag{*}
qhP - qhPQ + O(qhP^2) \leq M_2(q; h) \leq qhP
$$
where $Q=\prod_{\substack{{p \mid q}\\{p>h}}} (1-1/p)$.*

However, the careful argument is omitted, and I haven't been able to lure out the first inequality myself. I hope asking here could help me in that direction. The introduction of Montgomery and Vaughan's paper is included below as background.

**Question:** How can one derive

$$
qhP - qhPQ + O(qhP^2) \leq M_2(q; h)
$$
from (2) below?

### Background

Let $q$ be a natural number, let $P = \phi(q)/q$ be the "probability" that a randomly chosen integer is relatively prime to q, and let $$ \tag{1} M_k(q;h) = \sum_{n=1}^{q} \left( \sum_{\substack{{m=1}\\{(m+n,q)=1}}}^{h} 1 - h P \right)^k. $$ This is the $k$-th moment of the number of reduced residues modulo $q$ in an interval of length $h$ about its mean, $hP$. Clearly $M_1(q; h) = 0$. By an elementary calculation (see Hausman and Shapiro [3]) it may be shown that $$\tag{2} M_2(q;h) = qP^2 \sum_{\substack{{r \mid q }\\{r > 1}}} \mu(r)^2 \left( \prod_{\substack{ {p \mid q }\\{p \nmid r} }} \frac{p(p-2)}{(p-1)^2} \right) r^2 \phi(r)^{-2} \left\{ \frac{h}{r}\right\}\left( 1 - \left\{ \frac{h}{r}\right\}\right). $$ This with the simple inequality $\{\alpha\}(1 - \{\alpha\}) \leq \alpha$ gives immediately the estimate $$\tag{3} M_2(q;h)\leq qhP. $$ With a more careful argument from (2) it is easily seen that $$ qhP - qhPQ + O(qhP^2) \leq M_2(q; h) \leq qhP $$ where $Q=\prod_{\substack{{p \mid q}\\{p>h}}} (1-1/p)$.