已知a＞b＞c，求证：1a−b+1b−c≥4a−c．

Question

已知a＞b＞c，求证：(1)/(a−b)+(1)/(b−c)≥(4)/(a−c)．

Answer 1

证明：∵(a−c)/(a−b)+(a−c)/(b−c)=(a−b+b−c)/(a−b)+(a−b+b−c)/(b−c)＝2+(b−c)/(a−b)+(a−b)/(b−c)≥2+2√((b−c)/(a−b)×(a−b)/(b−c))=4，（a＞b＞c）
∴(a−c)/(a−b)+(a−c)/(b−c)≥4
∴(1)/(a−b)+(1)/(b−c)≥(4)/(a−c)