Since it holds that $\frac{1}{a-1} - \frac{8}{b+c} = \frac{1}{a-1} - \frac{8}{4-a} = \frac{12-9a}{(a-1)(4-a)} = \frac{3(4-3a)}{(a-1)(4-a)}$ the given inequality is equivalent to
$$3 \left( \frac{4-3a}{(a-1)(4-a)} + \frac{4-3b}{(b-1)(4-b)} + \frac{4-3c}{(c-1)(4-c)} \right) \ge 0.$$
Without loss of generality we can assume that $a \ge b \ge c$. Then clearly it holds $4-3a \le 4-3b \le 4-3c$. From $1 < a, b, c < 4$ it follows that $\frac{1}{(a-1)(4-a)}$, $\frac{1}{(b-1)(4-b)}$, $\frac{1}{(c-1)(4-c)}$ are positive real numbers. We will prove that $(a-1)(4-a) \ge (b-1)(4-b)$.
We have $(a-1)(4-a) \ge (b-1)(4-b) \Leftrightarrow 5a - a^2 \ge 5b - b^2 \Leftrightarrow (a-b)(5-a-b) \ge 0$. Analogously $(b-1)(4-b) \ge (c-1)(4-c)$. Hence $\frac{1}{(a-1)(4-a)} \le \frac{1}{(b-1)(4-b)} \le \frac{1}{(c-1)(4-c)}$. Since $4-3a \le 4-3b \le 4-3c$ we can use Chebyshev's inequality to obtain:
$$\frac{4-3a}{(a-1)(4-a)} + \frac{4-3b}{(b-1)(4-b)} + \frac{4-3c}{(c-1)(4-c)} \ge \frac{4-3a+4-3b+4-3c}{3} \left( \frac{1}{(a-1)(4-a)} + \frac{1}{(b-1)(4-b)} + \frac{1}{(c-1)(4-c)} \right) = 0.$$
Equality holds for $4-3a = 4-3b = 4-3c$ i.e. $a = b = c = \frac{4}{3}$.

Решение:
Бидејќи важи $\frac{1}{a-1} - \frac{8}{b+c} = \frac{1}{a-1} - \frac{8}{4-a} = \frac{12-9a}{(a-1)(4-a)} = \frac{3(4-3a)}{(a-1)(4-a)}$ даденото неравенство е еквивалентно со
$$3 \left( \frac{4-3a}{(a-1)(4-a)} + \frac{4-3b}{(b-1)(4-b)} + \frac{4-3c}{(c-1)(4-c)} \right) \ge 0.$$
Без губење на општоста нека претпоставиме дека $a \ge b \ge c$. Тогаш јасно е дека важи $4-3a \le 4-3b \le 4-3c$. Од $1 < a, b, c < 4$ следува $\frac{1}{(a-1)(4-a)}$, $\frac{1}{(b-1)(4-b)}$, $\frac{1}{(c-1)(4-c)}$ се позитивни реални броеви. Ќе докажеме дека $(a-1)(4-a) \ge (b-1)(4-b)$.
Имаме $(a-1)(4-a) \ge (b-1)(4-b) \Leftrightarrow 5a - a^2 \ge 5b - b^2 \Leftrightarrow (a-b)(5-a-b) \ge 0$. Аналогно $(b-1)(4-b) \ge (c-1)(4-c)$. Оттука следува дека $\frac{1}{(a-1)(4-a)} \le \frac{1}{(b-1)(4-b)} \le \frac{1}{(c-1)(4-c)}$. Бидејќи $4-3a \le 4-3b \le 4-3c$ можеме да го искористиме неравенство на Чебишев и добиваме:
$$\frac{4-3a}{(a-1)(4-a)} + \frac{4-3b}{(b-1)(4-b)} + \frac{4-3c}{(c-1)(4-c)} \ge \frac{4-3a+4-3b+4-3c}{3} \left( \frac{1}{(a-1)(4-a)} + \frac{1}{(b-1)(4-b)} + \frac{1}{(c-1)(4-c)} \right) = 0.$$
Равенство важи за $4-3a = 4-3b = 4-3c$ т.е. $a = b = c = \frac{4}{3}$.