Question

用综合法或分析法证明：已知a＞b＞c，求证〔1/a-b〕+〔1/b-c〕+〔1/c-a〕＞0

Answer 1

1/(a-b) + 1/(b-c) + 1/(c-a) > 0<== [(a-b)(b-c) + (b-c)*(c-a) + (c-a)*(a-b)] / [(a-b)(b-c)(c-a) > 0<== [ab- ac - b^2 + bc + bc - ab - c^2 + ac + ac -bc - a^2 + ab] / [(a-b)(b-c)(c-a)] > 0<== [ab + bc + ac - a^2 - b^2 - c^2] / [(a-b)(b-c)(c-a)] > 0<== [ab + bc + ac - a^2 - b^2 - c^2] < 0<== a^2 + b^2 + c^2 - ab - bc - ac > 0<== (a - b)^2 + (b - c)^2 + (c - a)^2 > 0 恒成立。