# The average number of large prime factors of $p-1$

How can one prove that $$\sum_{p \leq x} \mathop{\sum_{q | p-1}}_{q > x^{1/3}} 1 \leq 3\pi(x),$$ where both sums run over primes?

The left-hand side is $\displaystyle{\sum_{x^{1/3} < q < x} \pi(x;q,1)}$, but that doesn't seem to lead anywhere.

Just a hint please.

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How many distinct $q> x^\frac{1}{3}$ can divide $p$? –  N. S. Oct 31 '11 at 2:44

$$\sum_{n{\rm\ in\ }S}f(n)\le(\max_{n{\rm\ in\ }S}f(n))({\rm cardinality\ of\ }S)$$
I'm having trouble seeing why $\max_{p\le x} \#\{q:q|(p-1),\; q>x^{1/3}\}\le3$, and based on the lack of votes I'd say I'm not alone... –  anon Oct 31 '11 at 4:15
@anon, remember, $q$ is running over primes. If primes $q_1,q_2,q_3$ all exceed $x^{1/3}$, can they all divide a number which is less than $x$? –  Gerry Myerson Oct 31 '11 at 5:45