A question regarding the prime factors of a number and the prime-counting function

Question

Consider the inequality: $$\sum_{p\;|\;n} \frac{1}{p}>1-{\pi(n) \over n}$$ Where $\pi(x)$ is prime-counting function, $p\;|\;n$ means $p$ is a prime factor of $n$, and $n$ is natural number. My question is: can we prove such a thing, and if so, how?

Answer 1

This is false. Since $\pi(n)/n\to 0$ as $n\to\infty$ (by the prime number theorem, say), the right-hand side gets arbitrarily close to $1$. But there are arbitrarily large values of $n$ (primes, for instance) for which the left-hand side stays a fixed distance below $1$.

Indeed, it fails already for $n=2$, where both sides of the inequality are $1/2$. (Or it fails for $n=1$ where the left side is $0$, but perhaps you don't want to include that case.) In fact, it fails for every prime value of $n$, since it is easy to see that $\pi(n)/n<1/2$ for all $n>8$ (since even numbers other than $2$ are never prime).