This question is part of a proof in course in Sieve Theory( http://www.math.tau.ac.il/~rudnick/courses/sieves2015.html, precisely lecture 11 and 14)and I am not able to prove this particular statement. I have thought about it enough but couldn't make progress so I am posting it.
Let $S(z) = \sum_{n\leq z , n |P(z)} \frac{\mu^2(n)} { (\mu * f)(n)}$. Define the multiplicative function w(n) such that $w(p^k) =0 $ for $k\geq 2$ and w(p)= { p/ f(p) if $p \in P$ , 0 if $p \notin P$} . I have proved that $S(z) = \sum_{ d \leq z} \frac { w(d) } { d \prod_{p | d} ( 1- w(p) /p)}$.
I am struck on the an assertion of proof of Linnik's theorem( on page 3 of lecture 14): For square frees $n = p_1 ...p_n$, $w(n) = w(p_1) ...w(p_n)$. A={1,...,N} , P = { $p \leq N^{1/2} : (n/p) =1 $for all $n \leq N^{\epsilon}$} and $\Omega_p = ${ h ( mod p) : (h/p) =-1}. w(p) = #$\Omega_p = (p-1)/2$, p>2.
Prove that $S(z) \geq \sum_{p\leq z} w(p)/p = 1/2 \sum_{p \in P} (1- 1/p)$.
Kindly help me with this!