已知a>b>c 求证 1/(a-b)+1/(b-c)+1/(c-a)>0
热心网友
∵abc∴a-ca-b0,b-c0, 1/(b-c)0 .(1)∴1/(a-b)1/(a-c)即1/(a-b)-1/(a-c)0即1/(a-b)+1/(c-a)0 .(2)由(1)+(2)得1/(a-b)+1/(b-c)+1/(c-a)0
热心网友
优化上有原题的嘛
热心网友
1/(a-b)+1/(b-c)+1/(c-a)=(a-c)/(a-b)(b-c)-1/(a-c)=[(a-c)^2-(a-b)(b-c)]/(a-b)(b-c)(a-c)0