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Thread: Inequality

  1. #1
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    Inequality

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

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

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

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

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

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