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Question

If a, b, c are three positive real number then show : $$\displaystyle \frac{a}{b\, +\, c}\, +\, \frac{b}{a\, +\, c}\, +\, \frac{c}{a\, +\, b}\, \geq\, \frac{3}{2}$$


Solution

$$\dfrac { \dfrac { a }{ b\, +\, c } \, +\, \dfrac { b }{ a\, +\, c } \, +\, \dfrac { c }{ a\, +\, b }  }{ 3 } \, \geq \, \dfrac { 3 }{ \dfrac { b\, +\, c }{ a } \, +\, \dfrac { a\, +\, c }{ b } \, +\, \dfrac { a\, +\, b }{ c }  } $$  ($$AM>HM$$)
 $$\dfrac { \dfrac { a }{ b\, +\, c } \, +\, \dfrac { b }{ a\, +\, c } \, +\, \dfrac { c }{ a\, +\, b }  }{ 3 } \, \geq \, \dfrac { 3 }{ \dfrac { b }{ a } \, +\, \dfrac { a }{ b } \, +\, \dfrac { c }{ a } \, +\, \dfrac { a }{ c } \, +\, \dfrac { c }{ b } \, +\, \dfrac { b }{ c }  } $$
 $$\dfrac { \dfrac { a }{ b\, +\, c } \, +\, \dfrac { b }{ a\, +\, c } \, +\, \dfrac { c }{ a\, +\, b }  }{ 3 } \, \geq \, \dfrac { 3 }{ 6 } \, \left( \because \, \dfrac { a }{ b } \, +\, \dfrac { b }{ a } \, \geq \, 2 \right) $$($$AM>GM$$)
$$\dfrac { a }{ b\, +\, c } \, +\, \dfrac { b }{ a\, +\, c } \, +\, \dfrac { c }{ a\, +\, b } \, \geq \, \dfrac { 3 }{ 2 } $$

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