a>b>c
有a-c>0,a-b>0,b-c>0
a-c=(a-b)+(b-c)
(a-c)[1/(a-b)+1/(b-c)]
=(a-c){[(a-b)+(b-c)]/[(a-b)(b-c)]}
=[(a-c)^2]/[(a-b)(b-c)]
={[(a-b)+(b-c)]^2}/[(a-b)(b-c)]
={[(a-b)^2]/[(a-b)(b-c)]}+{[(b-c)^2]/[(a-b)(b-c)]}+2
=[(a-b)/(b-c)]+[(b-c)/(a-b)]+2
因为a-b>0,b-c>0
[(a-b)/(b-c)]+[(b-c)/(a-b)]
>=2√{[(a-b)/(b-c)]+[(b-c)/(a-b)]}
=2
[(a-b)/(b-c)]+[(b-c)/(a-b)]+2>=4
(a-c)[1/(a-b)+1/(b-c)]>=k恒成
所以4>=k
k最大值为4