Applications of Bounds of the Sum of Inverse Prime Divisors $\sum_{p \mid n} \frac{1}{p}$ Question: What interesting or notable problems involve the additive function $\sum_{p \mid n} \frac{1}{p}$ and depend on sharp bounds of this function? 
I'm aware of at least one: A certain upper bound of the sum implies the ABC conjecture.
By the Arithmetic-Geometric Inequality, one can bound the radical function (squarefree kernel) of an integer,
\begin{align}
\left( \frac{\omega(n)}{\sum_{p \mid n} \frac{1}{p}} \right)^{\omega(n)} \leqslant \text{rad}(n).
\end{align}
If for any $\epsilon > 0$ there exists a finite constant $K_{\epsilon}$ such that for any triple $(a,b,c)$ of coprime positive integers, where $c = a + b$, one has
\begin{align}
\sum_{p \mid abc} \frac{1}{p} < \omega(abc) \left( \frac{K_\epsilon}{c} \right)^{1/((1+\epsilon) \omega(abc))},
\end{align}
then
\begin{align}
c < K_{\epsilon} \left( \frac{\omega(abc)}{\sum_{p \mid abc} \frac{1}{p}} \right)^{(1+ \epsilon)\omega(abc)} \leqslant \text{rad}(abc)^{1+\epsilon},
\end{align}
and the ABC-conjecture is true.
Edit: Now, whether or not any triples satisfy the bound on inverse primes is a separate issue. Greg Martin points out that there are infinitely many triples which indeed violate it. This begs the question of whether there are any further refinements of the arithmetic-geometric inequality which remove such anomalies, but this question is secondary.

As it turns out, your proposed inequality is false - in fact false for any $\epsilon>0$, even huge $\epsilon$.
The following approximation to the twin primes conjecture was proved by Chen's method: there exist infinitely many primes $b$ such that $c=b+2$ is either prime or the product of two primes. With $a=2$, this gives $\omega(abc)\le4$, and so
$$
\omega(abc) \left( \frac{K_\epsilon}{c} \right)^{1/((1+\epsilon) \omega(abc))} \le 4 \bigg( \frac{K_\epsilon}c \bigg)^{1/(4+4\epsilon)},
$$
which (for any fixed $\epsilon>0$) can be arbitrarily small as $c$ increases. However, the other side $\sum_{p\mid abc} \frac1p$ is at least $\frac12$, and so the inequality has infinitely many counterexamples.
