Cho a, b, c dương thỏa mãn ab + bc + ca = 1. Chứng minh rằng:

Question

Cho a, b, c dương thỏa mãn ab + bc + ca = 1. Chứng minh rằng:

cho-a-b-c-duong-thoa-man-ab-bc-ca-1-chung-minh-rang

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Doris 10 months 2020-11-10T18:04:32+00:00 1 Answers 112 views 0

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    2020-11-10T18:05:57+00:00

    Ta có:

    $\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca}$

    $= \dfrac{ab+ bc+ ca}{ab} + \dfrac{ab+ bc+ ca}{bc} + \dfrac{ab+ bc+ ca}{ca}$

    $= 1 + \dfrac{bc + ca}{ab} + 1 + \dfrac{ab + ca}{bc} + 1 + \dfrac{ab + bc}{ca}$

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

    Mặt khác:

    $\sqrt{\dfrac{(a + b)(a + c)}{a^2}} \leq \dfrac{a +b + a + c}{2a} = \dfrac{2a + b + c}{2a} = 1 + \dfrac{1}{2}\left(\dfrac{b}{a} + \dfrac{c}{a}\right)$

    Tương tự:

    $\sqrt{\dfrac{(b + c)(b + a)}{b^2}} \leq 1 + \dfrac{1}{2}\left(\dfrac{c}{b} + \dfrac{a}{b}\right)$

    $\sqrt{\dfrac{(c + a)(c+ b)}{c^2}} \leq 1 + \dfrac{1}{2}\left(\dfrac{a}{c} + \dfrac{b}{c}\right)$

    $\Rightarrow 3 + \sqrt{\dfrac{(a + b)(a + c)}{a^2}} + \sqrt{\dfrac{(b + c)(b + a)}{b^2}} \sqrt{\dfrac{(c + a)(c+ b)}{c^2}} \leq 6 + \dfrac{1}{2}\left(\dfrac{c}{a} + \dfrac{c}{b} + \dfrac{a}{c} + \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{b}{a}\right)$

    Bất đẳng thức đã cho trở thành:

    $3 + \dfrac{c}{a} + \dfrac{c}{b} + \dfrac{a}{c} + \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{b}{a} \geq 6 + \dfrac{1}{2}\left(\dfrac{c}{a} + \dfrac{c}{b} + \dfrac{a}{c} + \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{b}{a}\right)$

    $\Leftrightarrow \dfrac{1}{2}\left(\dfrac{c}{a} + \dfrac{c}{b} + \dfrac{a}{c} + \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{b}{a}\right) \geq 3$ $(*)$

    Áp dụng bất đẳng thức $AM-GM$ ta được:

    $\dfrac{1}{2}\left(\dfrac{c}{a} + \dfrac{c}{b} + \dfrac{a}{c} + \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{b}{a}\right) \geq \dfrac{1}{2}.6\sqrt[6]{\dfrac{c}{a}\cdot\dfrac{c}{b}\cdot\dfrac{a}{c}\cdot\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{b}{a}} = \dfrac{1}{2}.6 = 3$

    $\Rightarrow (*)$ đúng

    Vậy bất đẳng thức được chứng minh

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