INEQUALITY'S BLOG

November 26, 2010

Problem 71: van khea

Filed under: the new math — KKKVVV @ 8:50 am

Let a, b, c be positive real numbers such that a\leq b\leq c. Prove that:
\displaystyle \frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq \sqrt{\frac{3}{2}(a+b+c)}

6 Comments »

  1. hello I was fortunate to search your Topics in yahoo
    your post is splendid
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    Comment by bet365 — December 15, 2010 @ 6:34 pm

  2. អូ! លោក​បង អា​ហ្នឹង​គេ​ប្រើ​ Cauchy-Schwarz មែន​ទេ?

    Comment by rainymathboy — December 25, 2010 @ 5:11 am

  3. លំហាត់នឹងដូចជាមិនប្រើ Cauchy – Schwarz ទេ។

    Comment by vankhea — December 25, 2010 @ 9:10 am

  4. ហ្នឹង​ហើយ ប្រើ​ហើយ​អត់​ចេញ​ផង…

    Comment by rainymathboy — December 25, 2010 @ 12:22 pm

  5. តែបើតាមខ្ញុំដឹងប្រហែលជាបានចូលមើលចំលើយហើយមើលទៅ ?? ហិហិ ម៉េចដែរក្បួតធ្វើលំហាត់របស់ខ្ញុំ ??

    Comment by khea — December 25, 2010 @ 6:47 pm


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