Question

已知a>b>c,求证：1/（a-b)+1/(b-c)+1/(c-a) >0

Answer 1

证 a>b>c

1/(a-b)+1/(b-c)+1/(c-a)
=(b-c)(c-a)+(a-b)(c-a)+(a-b)(b-c)/(a-b)(b-c)(c-a)
=bc-ba-c²+ca+ac-a²-bc+ab+ab-ac-b²+bc/(a-b)(b-c)(c-a)
=-(a²-ab+b²-ac+c²-bc)/-(a-b)(b-c)(a-c)
=-2(a²-ab+b²-ac+c²-bc)/-2(a-b)(b-c)(a-c)
=2a²-2ab+2b²-2ac+2c²-2bc/2(a-b)(b-c)(a-c)
=(a-b)²+(b-c)²+(a-c)²/2(a-b)(b-c)(a-c)

因为 a>b>c

所以 a-b>0 b-c>0 a-c>0
所以 2(a-b)(b-c)(a-c)>0

又因为 (a-b)²+(b-c)²+(a-c)²>0
所以 (a-b)²+(b-c)²+(a-c)²/2(a-b)(b-c)(a-c)>0

所以 1/（a-b)+1/(b-c)+1/(c-a)>0

Answer 2

1/(a-b)+1/(b-c)+1/(c-a)=1/(a-b)+(b-a)/(b-c)(c-a)=
[ab+bc+ac-a^2-b^2-c^2]/(a-b)(b-c)(c-a)=
-2[a^2+b^2+c^2-ab-bc-ac]/2(a-b)(b-c)(c-a)=
[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)]/2(a-b)(b-c)(a-c)=
[(a-b)^2+(b-c)^2+(a-c)^2]/2(a-b)(b-c)(a-c)＞0
即1/(a-b)+1/(b-c)+1/(c-a)>0