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Math Help - Inequality

  1. #1
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    Inequality

    Let a,b,c be positive real numbers. Prove that
    \dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\geq\d  frac{3}{2}
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  2. #2
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    Quote Originally Posted by math sucks View Post
    Let a,b,c be positive real numbers. Prove that
    \dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\geq\d  frac{3}{2}
    This happens to be well-known.

    \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}

    \frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a  +b} \geq \frac{9}{2}.

    \Delta = (a+b+c)\left( \frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b} \right)\geq \frac{9}{2}

    But, \frac{3}{\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b} } \leq \frac{2(a+b+c)}{3} by HM-AM.

    Thus, \Delta \geq (a+b+c)\cdot \frac{9}{2(a+b+c)} = \frac{9}{2}.
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