A) 1
B) \[\frac{3}{2}\]
C) 2
D) \[\frac{5}{2}\]
Correct Answer: B
Solution :
Using \[AM\ge GM\] \[\frac{\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}{3}\ge \sqrt{\frac{abc}{(a+b)(b+c)(c+a)}}\] ?(i) Again using \[AM\ge GM\] \[\frac{a+b}{2}\ge \sqrt{ab},\frac{b+c}{2}\ge \sqrt{bc}\frac{c+a}{2}\ge \sqrt{ca}\] \[\Rightarrow \]\[(a+b)(b+c)(c+a)\ge 8abc\] \[\Rightarrow \]\[\sqrt[3]{\frac{abc}{(a+b)(b+c)(c+a)}}\le \frac{1}{2}\] \[\therefore \]From Eq. (i) \[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\le \frac{3}{2}\]
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