Macedonian Mathematical Olympiad

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 $(a-1)(4-a)\ge (b-1)(4-b)\iff 5a-a^2\ge 5b-b^2\iff (a-b)(5-a-b)\ge 0$. 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 $(a-1)(4-a)\ge (b-1)(4-b)\iff 5a-a^2\ge 5b-b^2\iff (a-b)(5-a-b)\ge 0$. 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. $a=b=c=\frac{4}{3}$.