"On the distribution of reduced residues" by Montgomery and Vaughan – missing careful argument wanted 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 $Q=\prod_{\substack{{p \mid q}\\{p>h}}} (1-1/p)$.
 A: 1. First we prove the upper bound in $(\ast)$. Using the original hint, and noting that $P=\phi(q)/q$, it suffices to show the identity
$$\sideset_{^\flat}\sum_{r\mid q}\frac{r}{\phi(r)^2}
\left(\prod_{\substack{ {p \mid q }\\{p \nmid r} }}\frac{p(p-2)}{(p-1)^2}
\right)=\frac{q}{\phi(q)},$$
where $\flat$ indicates that the summation is restricted to square-free values of $r$. The two sides are multiplicative in $q$, hence it suffices to verify the special case when $q$ is the power of a prime $p$. In that case, the identity boils down to
$$\frac{p(p-2)}{(p-1)^2}+\frac{p}{(p-1)^2}=\frac{p}{p-1},$$
which is evident.
2. Now we prove the lower bound in $(\ast)$, which can be rewritten as
$$\frac{M_2(q;h)}{qhP}\geq 1-Q+O(P).$$
Equivalently,
$$\frac{1}{\phi(q)}\sideset_{^\flat}\sum_{r\mid q}\frac{r}{h}\left\{ \frac{h}{r}\right\}\left( 1 - \left\{ \frac{h}{r}\right\}\right)\prod_{\substack{ {p \mid q }\\{p \nmid r} }}(p-2)\geq 1-Q+O(P).$$
It is clear that (cf. previous point)
$$\frac{1}{\phi(q)}\sideset_{^\flat}\sum_{r\mid q}\prod_{\substack{ {p \mid q }\\{p \nmid r} }}(p-2)=1,$$
hence the lower bound in $(\ast)$ is equivalent to
$$\frac{1}{\phi(q)}\sideset_{^\flat}\sum_{r\mid q}f(h,r)\prod_{\substack{ {p \mid q }\\{p \nmid r} }}(p-2)\leq Q+O(P),$$
where $f(h,r)$ abbreviates
$$f(h,r):=1-\frac{r}{h}\left\{ \frac{h}{r}\right\}\left( 1 - \left\{ \frac{h}{r}\right\}\right).$$
It is straightforward that
$$f(h,r)\leq\min\left(1,\frac{h}{r}\right)\leq\prod_{\substack{p\mid r\\p>h}}\frac{h}{p},$$
hence hence it suffices that
$$\frac{1}{\phi(q)}\left(\prod_{\substack{p\mid q\\p\leq h}}(p-2+1)\right)
\left(\prod_{\substack{p\mid q\\p>h}}\left(p-2+\frac{h}{p}\right)\right)
\leq Q+O(P).$$
Equivalently,
$$\prod_{\substack{p\mid q\\p>h}}\left(1-\frac{1}{p-1}+\frac{h}{p(p-1)}\right)\leq Q+O(P).$$
Now the left hand side equals
$$Q\prod_{\substack{p\mid q\\p>h}}e^{O(h/p^2)}=Q\left(1+\frac{O(1)}{\log h}\right)=Q+O\left(\frac{Q}{\log h}\right)=Q+O(P),$$
and we are done. In the last step, we used that
$$Q=P\prod_{\substack{p\mid q\\p\leq h}}\left(1-\frac{1}{p}\right)^{-1}=O(P\log h).$$
