mycroft holmes
Last Activity: 9 Years ago
If a2, b2, c2 are in A.P., so are a2+ab+bc+ca, b2+ab+bc+ca, and c2+ab+bc+ca
i.e. (a+b)(c+a), (b+c)(a+b), (c+a)(b+c) are in A.P.
Multiplying the three terms by (a+b+c)/ (a+b)(b+c)(c+a), we have
(a+b+c)/(b+c), (a+b+c)/c+a, (a+b+c)/a+b are in A.P. Now subtracting 1 from each term, we have
1/b+c, 1/c+a, 1/a+b are in A.P.