Cho `a>b>0.CM: (a+b)/2 < (a-b)^2/(8b)+ \sqrt{ab}`

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Cho `a>b>0.CM: (a+b)/2 < (a-b)^2/(8b)+ \sqrt{ab}`

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Hưng Khoa 8 tháng 2020-10-31T17:17:46+00:00 1 Answers 64 views 0

Answers ( )

  1. Đáp án:

    $\frac{a+b}{2} <\frac{(a-b)^2}{8b} +\sqrt{ab}$ 

    Giải thích các bước giải:

    $\frac{a+b}{2} <\frac{(a-b)^2}{8b} +\sqrt{ab}$ 

    $⇔a+b < \frac{(a-b)^2}{4b}+2\sqrt{ab}$ 

    $⇔a-2\sqrt{ab} +b< \frac{(a-b)^2}{4b}$

    $⇔(\sqrt{a}-\sqrt{b})^2.4b< (a-b)^2$

    $⇔(\sqrt{a}-\sqrt{b})^2.4b<[ (\sqrt{a}-\sqrt{b}).(\sqrt{a}+\sqrt{b})]^2$

    $⇔4b <(\sqrt{a}+\sqrt{b})^2$

    $⇔0 <(\sqrt{a}+\sqrt{b})^2-(2\sqrt{b})^2$

    $⇔0< (\sqrt{a}+\sqrt{b}-2\sqrt{b}).(\sqrt{a}+\sqrt{b}+2\sqrt{b})$

    $⇔0<(\sqrt{a}-\sqrt{b}).(\sqrt{a}+3\sqrt{b})$

    vì a >b > 0 nên$ \sqrt{a}-\sqrt{b}>0 ;(\sqrt{a}+3\sqrt{b})>0$

    ⇒$(\sqrt{a}-\sqrt{b}).(\sqrt{a}+3\sqrt{b}) >0$

    ⇒$\frac{a+b}{2} <\frac{(a-b)^2}{8b} +\sqrt{ab}$ 

               _đpcm_

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