等差数列求证问题

(b+c)/a+(a+b)/c
=(1+(b+c)/a)+(1+(a+b)/c)-2
=(a+b+c)/a+(a+b+c)/c-2
=(a+b+c)(1/a+1/c)-2
=(a+b+c)*2/b-2
=2(a+c)/b+2b/b-2
=2(a+c)/b+2-2
=2(a+c)/b
所以：（b＋c）/a，（c＋a）/b，（a＋b）/c也成等差数列

因为1/a,1/b,1/c是等差数列，所以2/b=1/a+1/c=(a+c)/ac
故a+c=2ac/b，ab+bc=2ac
(b+c)/a+(a+b)/c-2(a+c)/b
=(bc+c^2+a^2+ab)/ac-4ac/b^2
=(b^3c+b^2c^2+a^2b^2+ab^3-4a^c^2)/acb^2
=(b^3c+b^2c^2+a^2b^2+ab^3-a^2b^2-b^2c^2-2ab^2c)/acb^2
=(b^3c+ab^3-2ab^2c)/acb^2
=(bc+ab-2ac)/ac
=0
所以(b+c)/a,(a+c)/b,(a+b)/c是成等差数列