What is the significance of $\prod \limits_{i=1}^{t}m_i > p \prod \limits_{i=1}^{t-1}m_{n-i+1}$ in Asmuth–Bloom threshold SSS?
One of the required properties with a secret sharing scheme is that if you have $t-1$ shares (where $t$ is the threshold), then you can't eliminate any possible values from being the shared secret.

Now, suppose that we had $t-1$ shares; with the Asmuth-Bloom scheme, we would learn the most if they happened to be shares $(y_{n-t+2})$ through $y_{n}$.  If you had those shares, then you could deduce the value $y \bmod \prod \limits_{i=1}^{t-1}m_{n-i+1}$

You also know that:


$0 \le y &lt; M$
You really want the value $y \bmod p$


For you to learn nothing, we need to have all possible values of $y \bmod \prod \limits_{i=1}^{t-1}m_{n-i+1} + pk$ to be in the range $[0, .., M-1]$

This is guaranteed if $p \cdot \prod \limits_{i=1}^{t-1}m_{n-i+1} \ge M$, which (if we remember that $M = \prod \limits_{i=1}^t m_i$ is precisely the restriction they give.

BTW: what is this fascination with Asmuth-Bloom?  In my opinion, Shamir Secret Sharing is a better way of solving the problem; it's simpler, a lot easier to set up, has smaller shares, and it doesn't leak any probabilistic information like Asmuth-Bloom does.