已知a>b>c>d,求证：1/（a-b)+1/(b-c)+1/(c-a)>=9/(a-d).
1个回答
你题目中把 (c-d) 误打成 c-a 了
只需证明 （a-d) [1/(a-b)+1/(b-c)+1/(c-d)] >= 9
[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)
= [1 + (b-c)/(a-b) + (c-d)/(a-b)] + [1 + (a-b)/(b-c) + (c-d)/(b-c)] + [1 + [(a-b)/(c-d) + (b-c)/(a-d)]
= 3 + [(b-c)/(a-b) + (a-b)/(b-c)] + [(c-d)/(a-b) + (a-b)/(c-d)] + [(c-d)/(b-c) + (b-c)/(c-d)]
≥ 3 + 2 + 2 + 2
= 9
所以 ：1/（a-b)+1/(b-c)+1/(c-d)>=9/(a-d)
题目中用到了（x-y)^2≥0 所以 x^2 + y^2 ≥2xy 这一性质
例如 [(b-c)/(a-b) + (a-b)/(b-c)] ≥ 2 √[(b-c)/(a-b)] * √[(a-b)/(b-c)] = 2
相关问题
