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.