Some simple and often used properties of the function $ s(n)$ are: $ s(a + b)\le s(a) + s(b)$. This is really easy to prove by analyzing the process of adding $ a$ and $ b$ 'in column'. The second property is $ s(mn)\le s(m)s(n)$. To see this let firstly note that for a digit $ 1\le d\le9$ by the first property we have $ s(md)\le ds(n)$. Now let $ n = \overline{a_1a_2\ldots a_k}$ be the decimal representation of $ n$. Then $ s(mn) = s(ma_110^{k - 1} + ma_210^{k - 2} + \ldots + ma_{k - 1}\cdot10 + ma_k)\le$ $ s(ma_110^{k - 1}) + s(ma_210^{k - 2}) + \ldots + s(ma_{k - 1}\cdot10) + s(ma_k) = E$. Further $ E = s(ma_1) + s(ma_2) + \ldots + s(ma_k)\le s(m)[a_1 + \ldots + a_k] = s(m)s(n)$ and we are done. Now the desired inequality immediately results from $ s(n) = s(1000n) = s(125\cdot8n)\le8s(8n)$.

wow, the above is elegant. I found two proofs based on induction, one of which I will give in full detail. Every letter will denote a positive integer. Assume we have $S(8n) \ge \frac{1}{8}S(n)$ for all $n \le 1.25 * 10^{k+2}$ for some $k \ge 1$ (it can be easily checked to hold for $k = 1$). Then, for every $a \le 10^k$, we have: $S(a) = S(1000a) = S(8*(125a)) \ge \frac{1}{8}S(125a) = \frac{1}{8}S(1250a)$ Now, set $n = 1250a + b$ with $a \le 10^k$, $b \le 1249$. We then have: $\mbox{S(8n) = S(10000a + 8b) = S(a) + S(8b) \ge \frac{1}{8}S(1250a) + \frac{1}{8}S(b) \ge \frac{1}{8}S(1250a + b) = \frac{1}{8}S(n)}$ And we see that our inequality holds for all $n \le 1250*10^k = 1.25*10^{k+3}$. This finishes our induction step. The second proof, while also based on induction, is completely different; it feels more like a direct computation. Let $k = k(n)$, $n = \sum_{j=0}^k A_{j,n}*10^j$ and let $n'$ be $n$ without its first digit; $n' = \sum_{j=0}^{k-1} A_{j,n}*10^j$. Now, we have the following (non-trivial) property that a smallest counterexample $> 125$ to the inequality $S(8n) \ge \frac{1}{8}S(n)$ must have: $100*A_{i+2,n} + 10*A_{i+1,n} + A_{i,n} > 125$ for all $0 \le i \le k - 2$. In essence this boils down to the fact that a counterexample that doesn't satisfy that property can be split into two smaller parts, at least one of which also has to be a counterexample. For example: if 9870654 is a counterexample (note that 100*0 + 10*6 + 5 = 65 < 125), either 987 or 654 must be a counterexample too. Now, assume that $S(8n) \ge \frac{1}{8}S(n) + A_{k,8n}$ holds for all $n < 10^m$ that satisfy $100*A_{i+2,n} + 10*A_{i+1,n} + A_{i,n} > 125$ for all $0 \le i \le k - 2$ and set $n$ between $10^m$ and $10^{m+1}$. We can then write $S(8n)$ in terms of $S(8n')$ and some digits of $8n$ and $8n'$. Now, use the induction hypothesis on $S(8n')$, use the fact that $S(n') = S(n) - A_{k,n}$ and you're basically done. If I'm not mistaken, it can be shown that $S(8n) \ge \frac{1}{8}S(n) + A_{k,8n}$ holds for all $n > 1$, which is (albeit very marginally) more than what was asked for. But it should be clear that the extra term $A_{k,8n}$ came with a price: a longer and uglier proof that has no way of being generalised and is nowhere near as powerful as the easy and beautiful inequality of freemind: $s(mn) \le s(m)s(n)$