We recall that [abc \geq 8(s-a)(s-b)(s-c)] for any triangle, which directly leads to [\fracabcs(s-a)(s-b)(s-c) \geq \frac8(s-a)(s-b)(s-c)s(s-a)(s-b)(s-c) = \frac8s.] However, a more straightforward path to $n \geq 1$ involves leveraging known inequalities that directly compare $abc$ and $s(s-a)(s-b)(s-c)$.
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