Let, $c>1$ be any arbitrary integer. Let it's prime factorization be expressed as:
$$c = \prod_{i=1} p_i $$
Then I can show
$$\sum_{i} p_i > \frac{\ln(c)}{e^{1/e}}$$
where $e$ is the exponential function. This made me wonder if there were better estimates available in the literature?
Example
$c = 18$ The primes are $2,3,3$ then the sum is $\sum_{i} p_i = 2+3+3=8 > \ln(18)/e^{1/e}$
This is more a fact about calculus than about the integers. Define the function
$$f(x)=\frac{\ln(x)}x$$
we have that $f'(x)=\frac{1-\ln(x)}{x^2}$ which is only zero at $x=e$ which we can verify is the global maximum so $f(x)\leq f(e)= \frac 1e$ for all $x$, or $\ln(x)\leq \frac xe$, so
$$\ln(c)=\ln(p_1)+\dots+\ln(p_n)$$
$$\ln(c)\leq \frac 1e\left(p_1+\dots +p_n\right)$$
$$p_1+\dots +p_n\geq e\ln(c)>\frac{\ln(c)}{e^{1/e}}.$$