高2数学不等式
求证1/(a-b)+1/(b-c)+1/(c-d)大于等于9/(a-d) (a>b>c>d)
我要详细的解题过程.谢谢!
参考答案:[1/(a-b)+1/(b-c)+1/(c-d)]*(a-d)
=[1/(a-b)+1/(b-c)+1/(c-d)]*(a-b+b-c+c-d)
=[a-b+b-c+c-d/(a-b)]+[a-b+b-c+c-d/(b-c)]+[a-b+b-c+c-d/(c-d)]
=3+(b-c/a-b)+(c-d/a-b)+(a-b/b-c)+(c-d/b-c)+(a-b/c-d)+(b-c/c-d)
=3+(b-c/a-b)+(a-b/b-c)+(c-d/a-b)+(a-b/c-d)+(b-c/c-d))+(c-d/b-c)
≥3+2+2+2=9
所以1/(a-b)+1/(b-c)+1/(c-d)大于等于9/(a-d) (a>b>c>d)