$\pi(n)$ is always more than the sum of the prime indices of the factors of composite $n \geq 12$ Let $n=p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} \geq 12$ be any composite integer.
Then it seems that this is true: $$\pi(n) > \sum_{i=1}^{k}{\pi(p_i)a_i}\ .$$
You get equality instead iff $n$ is prime.
I also assume that if it is true, it's a known result. Can anyone point me towards a resource discussing it if so? Alternatively, if I've made a mistake and/or this is a trivial result, please point out how.

Also, if true, I think Bertrand's Postulate immediately follows:
For prime $p$, it gives $\pi(2p)>\pi(p)+1$, implying at least one prime between $p$ and $2p$.
By the same token, $\pi(3p)>\pi(p)+2$, and $\pi(p^2)>2 \pi(p)$.

(See my answer below for additional thoughts.)
 A: Not quite complete answer:
Suppose $km=n$ with $6\le k\le m$. A result of Rosser–Schoenfeld says that $\pi(x) < 1.25506x/\log x$ for $x>1$, so
$$
\pi(k)+\pi(m) \le 2\pi(m) < 2.51012m/\log m < 5.02024m/\log n,
$$
since $m\ge\sqrt n$. On the other hand, another result of Rosser–Schoenfeld says that $\pi(x) > x/\log x$ for $x>17$. The fact that $n\ge17$ and $k\ge6$ now forces
$$
\pi(n) > n/\log n = km>\log n > 5.02024m/\log n > \pi(k)+\pi(m).
$$
On the other hand, a result of Ramanujan says that $\pi(2x) \ge \pi(x) + 2$ and $\pi(3x) \ge \pi(x) + 3$ when $x\ge6$. Therefore when $m\ge6$,
\begin{align*}
\pi(2m) &> \pi(m) + 1 = \pi(m) + \pi(2) \\
\pi(4m) \ge \pi(3m) &> \pi(m) + 2 = \pi(2m) + \pi(3) \\
\pi(5m) &\ge \pi(4m) \ge \pi(2m)+2 > (\pi(m)+1)+2 = \pi(m) + \pi(5).
\end{align*}
In other words, we have shown that $\pi(k) + \pi(m) < \pi(km)$ for $k\ge2$ and $m\ge6$.
This should be very close to proving the entire statement by induction on the number of prime factors (counting multiplicity).
A: As shorthand, let $\sum_\pi(n)$ denote the sum of prime indices of $n$ as described above.
For any $k$, we know $\pi(p_k)=\sum_\pi(p_k)=k$.
From Bertrand's postulate, we know that $\pi(2^k) \geq \sum_\pi(2^k)=k$.
Any intermediate $n$ such that $\sum_\pi(n)=k$ will fall in the range $(p_k,2^k)$ and thus have $\pi(n) \geq k$. This can be shown explicitly by starting with $2^k$ and repeatedly dividing by $2$ while increasing another factor to its next larger prime. Bertrand ensures that the next larger prime will always be less than the factor of $2$ we lost, and so the overall product must also decrease.
e.g. $(1,1,1,1,1)\rightarrow (2,1,1,1)\rightarrow (2,2,1)\rightarrow (3,2)$, using integer partition tuples to represent the products $2^5=32$, $3\cdot 2^3=24$, $3^2\cdot 2=18$, $5\cdot 3=15$, and the minimum we could have gone to here is $(5)=p_5=11$.
Thus we see that any $n$ where $\sum_\pi(n)=k$ must be in the range $p_k \leq n \leq 2^k$, and since $\pi$ is a monotonically increasing function, we know that any such intermediate $n$ will give $\pi(p_k) \leq \pi(n) \leq \pi(2^k)$. And again, since $\sum_\pi(p_k)=k$, any $n>p_k$ will have $\pi(n)\geq k$.
This should suffice to prove $\pi(n) \geq \sum_\pi(n)$ for all $n\in\mathbb N$, or equivalently, that $\pi(ab)\geq \pi(a)+\pi(b)$.
To get the bounded inequality given in the problem, I think it's a matter of how many special cases you want to address; the larger the lower bound for $n$, the larger the constant you can reliably add to the right side of this, i.e. for $n\geq 12$, you can use $\pi(n) \geq \sum_\pi(n)+1$. In other words, for any $c$, there's some $N$ such that for all $n \geq N$, you get $\pi(n) \geq \sum_\pi(n)+c$.
