See here: http://www.artofproblemsolving.com/community/c6h1083654

Tintarn： $\frac {a+1} {a-1}>0,\frac {b+1} {b-1}>0,\frac {c+1} {c-1}>0,\frac {d+1} {d-1}>0 $ and $(a+1)(b+1)(c+1)(d+1)=(a-1)(b-1)(c-1)(d-1),$ $\frac {1} {a-1}+\frac {1} {b-1}+ \frac {1} {c-1}+ \frac {1} {d-1}$ $=\frac{1}{2}\left(\frac {a+1} {a-1}+\frac {b+1} {b-1}+\frac {c+1} {c-1}+\frac {d+1} {d-1}\right)-2$ $>2\sqrt[4]{ \frac {a+1} {a-1}\cdot \frac {b+1} {b-1}\cdot\frac {c+1} {c-1}\cdot\frac {d+1} {d-1}}-2=0.$

Let $ a,b,c$ be real numbers satisfying $|a|,|b|,|c|>1$ and $ab+bc+ca+1=0$. Prove that$$\frac {1} {a-1}+\frac {1} {b-1}+ \frac {1} {c-1}>0.$$

$(a + 1)(b + 1)(c + 1) = (a - 1)(b - 1)(c - 1)$ yields $$\frac{2}{a-1} + \frac{2}{b-1} + \frac{2}{c-1} = \frac{a+1}{a-1} + \frac{b+1}{b-1} + \frac{c+1}{c-1} - 3 > 3 - 3 = 0$$ by AM-GM (because $\frac{a+1}{a-1} = \frac{a^2 - 1}{(a-1)^2}$ is clearly positive and similarly for the others). Equality cannot hold because $a + 1 = a - 1$ is never true.

A slightly different solution using polynomials: Let $a,b,c,d$ be the roots of some polynomial $P$, so the quantity we are looking for is $-\frac{P'(1)}{P(1)}$. WLOG $P(1)$ is positive and from $abc+bcd+cda+dab+a+b+c+d = 0$, $P(-1) = P(1)$. Since $P$ has $4$ real roots and they must all be outside the range $[-1, 1]$, it is clear that $P'(1)$ is negative, so the desired quantity must be positive.

See also here http://problems.ru/view_problem_details_new.php?id=65255 http://www.artofproblemsolving.com/community/c6h1083654p4776641

This is also India IMOTC 2016, Practice test 1, Problem 3.

The relation is equivalent to \[(a + 1)(b + 1)(c + 1)(d + 1) = (a - 1)(b - 1)(c - 1)(d - 1).\]Since $a, b, c, d \notin [-1, 1]$, the fractions $\tfrac{a + 1}{a - 1}, \tfrac{b + 1}{b - 1}, \tfrac{c + 1}{c - 1}, \tfrac{d + 1}{d - 1}$ are positive, so AM-GM gives \[\frac{a + 1}{a - 1} + \frac{b + 1}{b - 1} + \frac{c + 1}{c - 1} + \frac{d + 1}{d - 1} \ge 4 \implies \frac{1}{a - 1} + \frac{1}{b - 1} + \frac{1}{c - 1} + \frac{1}{d - 1} \ge 0.\]Equality holds iff $\tfrac{a + 1}{a - 1} = \tfrac{b + 1}{b - 1} = \tfrac{c + 1}{c - 1} = \tfrac{d + 1}{d - 1} \iff a = b = c = d$, which is not possible.

All Russian Mathematics Olympiad problem.

um we already know that's the source of the problem --- it's in the title of the post

By conditions numbers $\frac{a+1}{a-1},\frac{b+1}{b-1},\frac{c+1}{c-1},\frac{d+1}{d-1}$ are positive and have product $1.$ Therefore $$\frac{2}{a-1}+\frac{2}{b-1}+\frac{2}{c-1}+\frac{2}{d-1}=\frac{a+1}{a-1}+\frac{b+1}{b-1}+\frac{c+1}{c-1}+\frac{d+1}{d-1}-4\stackrel{\text{AM-GM}}{\geq} 4-4=0,$$where equality holds only for $a=b=c=d,$ which is not true. The conclusion follows.