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

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

Suggest Corrections

0

Similar questions
View More

People also searched for
View More