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Kaya Renger Coordinator

15/08/2017 at 22:45
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Given a,b,c are positive real numbers so that abc = 1

Prove that : \(\dfrac{b+c}{\sqrt{a}}+\dfrac{c+a}{\sqrt{b}}+\dfrac{a+b}{\sqrt{c}}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)

Good luck :) 




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    AL 16/08/2017 at 15:58

    Use AM-GM's ineq:

    \(\dfrac{b+c}{\sqrt{a}}+\dfrac{c+a}{\sqrt{b}}+\dfrac{a+b}{\sqrt{c}}\ge\dfrac{2\sqrt{bc}}{\sqrt{a}}+\dfrac{2\sqrt{ca}}{\sqrt{b}}+\dfrac{2\sqrt{ab}}{\sqrt{c}}\)

    \(=\left(\sqrt{\dfrac{bc}{a}}+\sqrt{\dfrac{ca}{b}}\right)+\left(\sqrt{\dfrac{ca}{b}}+\sqrt{\dfrac{ab}{c}}\right)+\left(\sqrt{\dfrac{ab}{c}}+\sqrt{\dfrac{bc}{a}}\right)\)

    \(\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{a}+\sqrt{b}+\sqrt{c}\)

    \(\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[3]{abc}\)\(=\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)

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