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Kayasari Ryuunosuke Coordinator

14/07/2017 at 13:38
Answers
3
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Prove Nesbitt inequality :

Give a,b,c are positive real numbers , then , we have :

\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)

P/s : Use AM - GM inequality




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  • ...
    Lãng Tử Hào Hoa 14/07/2017 at 16:00

    We have BĐT equivalent:

    \(\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{a+b}+1\right)\) \(\ge\dfrac{9}{2}\)

    \(\Leftrightarrow\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge9\)

    Use AM - GM inequality we have:

    \(\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

    \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{3}{\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

    \(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\) (Thing must prove)

    The equality occurs \(\Leftrightarrow a=b=c\)

     
    Kayasari Ryuunosuke selected this answer.
  • ...
    Ace Legona 14/07/2017 at 15:33

    Another way - Use Cauchy-Schwarz

    \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)

    \(\Leftrightarrow\dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1+\dfrac{c}{a+b}+1\ge\dfrac{9}{2}\)

    \(\Leftrightarrow\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}\ge\dfrac{9}{2}\)

    \(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\ge\dfrac{9}{2}\)

    By Cauchy-Schwarz: \(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\ge\dfrac{9}{2\left(a+b+c\right)}\)

    \(\Rightarrow L.H.S=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\)

    \(\ge a+b+c\cdot\dfrac{9}{2\left(a+b+c\right)}=\dfrac{9}{2}=R.H.S\)

    The equality occurs for \(a=b=c\)

  • ...
    Phan Thanh Tinh Coordinator 14/07/2017 at 13:55

    Denote \(x=b+c;y=a+c;z=a+b\),then \(x,y,z>0\)

    Consider the following expression :

    \(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1+\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{y}{x}+1+\dfrac{y}{z}+\dfrac{z}{x}+\dfrac{z}{y}+1\)\(=3+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{x}{z}+\dfrac{z}{x}\right)+\left(\dfrac{y}{z}+\dfrac{z}{y}\right)\)

    We have : \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\sqrt{\dfrac{x}{y}.\dfrac{y}{x}}=2\) (AM-GM inequality)

    Similarly,\(\dfrac{x}{z}+\dfrac{z}{x}\ge2;\dfrac{y}{z}+\dfrac{z}{y}\ge2\).So,we have :

    \(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge3+2+2+2=9\)

    \(\Leftrightarrow2\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)\ge9\)

    \(\Leftrightarrow2\left(\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}\right)\ge9\)

    \(\Leftrightarrow1+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}+1+\dfrac{c}{a+b}\ge\dfrac{9}{2}\)

    \(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)


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