Cứu với chuyên gia, hsg cm:

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Cứu với chuyên gia, hsg cm:
cuu-voi-chuyen-gia-hsg-cm

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Acacia 6 years 2020-11-24T11:27:57+00:00 1 Answers 78 views 0

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  1. Doris
    0
    2020-11-24T11:29:11+00:00
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    Áp dụng bất đẳng thức $Schwarz$ ta được:

    $\dfrac{a + b + c}{3} = \dfrac{a}{3} + \dfrac{b}{3} + \dfrac{c}{3}$

    $\geq \dfrac{(\sqrt a + \sqrt b + \sqrt c)^2}{3 +3 + 3} = \left(\dfrac{\sqrt a + \sqrt b + \sqrt c}{3}\right)^2$

    $\geq \left[\left(\dfrac{\sqrt[4]{a} + \sqrt[4]{b} + \sqrt[4]{c}}{3}\right)^2\right]^2 = \left(\dfrac{\sqrt[4]{a} + \sqrt[4]{b} + \sqrt[4]{c}}{3}\right)^4$

    $\geq \left[\left(\dfrac{\sqrt[8]{a} + \sqrt[8]{b} + \sqrt[8]{c}}{3}\right)^2\right]^4 = \left(\dfrac{\sqrt[8]{a} + \sqrt[8]{b} + \sqrt[8]{c}}{3}\right)^8$

    $\Rightarrow \begin{cases}\sqrt[4]{\dfrac{a + b + c}{3}}\geq \dfrac{\sqrt[4]{a} + \sqrt[4]{b} + \sqrt[4]{c}}{3}\\\sqrt[8]{\dfrac{a + b + c}{3}}\geq \dfrac{\sqrt[8]{a} + \sqrt[8]{b} + \sqrt[8]{c}}{3}\end{cases}$

    Dấu = xảy ra $\Leftrightarrow a = b = c$

     

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