Sum over summation of distinct primes formula?

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Background

I found an interesting unorthodox proof of the below. Let:

$$S(x) = \text{Sum of all distinct primes of $x$}$$

and $ S(1) = 0$

Then:

$$ \sum_{r=1}^L S(r) = \sum_{i=1}^\infty p_i \Big[ \frac{L}{p_i} \Big] \leq\pi(L) L $$

where $p_i$ is the $i$'th prime and $[y]$ is the greatest integer function, $\pi(y)$ is the number of primes less than or equal to $y$

Question

What would a normal proof of the same look like? Is there a better estimate (asymptotic allowed)?

Example

$$ \sum_{r=1}^6 S(r) = S(2) +S(3) +S(4) +S(5) +S(6) = 2+3+2+5+5 = 17$$

$$ \sum_{r=1}^6 S(r) =2 \Big[\frac{6}{2} \Big] + 3 \Big[\frac{6}{3} \Big]+ 5 \Big[\frac{6}{5} \Big ] = 17 \leq 18 $$