Proof of inequality involving multiplicative function?

Question

The identity below seems true for the examples I've considered. I thought I had proven it using induction but found a mistake and removed my attempted proof since it is not helpful. Given:

$P(z)$ is the product of primes less than $z,$

$\nu(d)$ is the number of distinct prime divisors of $d$,

$\omega(\cdot)\geq 0$ is multiplicative, $m$ is an integer $\geq 1,$

can we prove/disprove

$\sum_{d|P(z),\nu(d) \leq m} \omega(d)\leq (1+\sum_{p|P(z)}\omega(p))^m\hspace{7mm}?$

Answer 1

Hint: Consider $\omega(p_1p_2...p_k)$, and how you could break this down using multiplicativity. Then see if this might match with a term (or combination of terms) on the right hand side.