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Problem #652

Chứng minh bất đẳng thức

Added by almost 3 years ago. Updated almost 3 years ago.

Status: Closed Start Date: 01-04-2010
Priority: Normal Due date: 09-04-2010
Assigned to: phuoclh % Done:

0%

Category: Bất đẳng thức
Target version: Xuân 2010
Votes: 2/2

Description

Cho \(a, b, c > 0\), chứng minh rằng

\[ \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} \geq \dfrac{3}{2} \]

History

Updated by tanphu almost 3 years ago

    Đây là bất đẳng thức Nesbit nổi tiếng, có nhiều cách chứng minh bất đẳng thức này, chủ yếu dựa vào bất đẳng thức Cauchy cho 3 số.

    Updated by almost 3 years ago

    • Due date set to 09-04-2010
    • Status changed from New to Assigned
    • Target version set to Xuân 2010

    Updated by themax_1311 almost 3 years ago

    • Votes: 1/1

    cái này dễ mà bạn

    Updated by almost 3 years ago

      DỄ thì max chứng minh đi
      sao nói xuông ko thế

      Updated by tanphu almost 3 years ago

      • Votes: 1/1

      Cách 1. Ta có
      \[
      \begin{array}{l l l}
      \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} &=& \dfrac{a}{b+c}+1 + \dfrac{b}{c+a}+1 + \dfrac{c}{a+b}+1 -3 \\
      &=& \dfrac{a+b+c}{b+c} + \dfrac{a+b+c}{c+a} + \dfrac{a+b+c}{a+b} -3 \\
      &=& (a+b+c)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right) -3 \\
      &=& \dfrac{1}{2}\Big[(b+c)+(c+a)+(a+b)\Big]\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right) -3 \\
      & \geq & \dfrac{9}{2}-3 \\
      &=&\dfrac{3}{2}
      \end{array}
      \]

      Updated by tanphu almost 3 years ago

      • Category set to Bất đẳng thức
      • Status changed from Assigned to Closed

      Updated by almost 3 years ago

      • Resolution set to Fixed