Question

If $a,b,c$ are positive real numbers, then $\begin{array}{rcl}& & \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\end{array}$ is

• A. $<\frac{3}{2}$
• B. $\ge \frac{3}{2}$
• C. $>\frac{5}{2}$
• D. None of these

Answer 1

The correct option is B

$\ge \frac{3}{2}$

Explanation for correct option.

Given, $a,b,c$ are positive real numbers

Now,

Arithmetic Mean of the term $\ge$inverse of the Arithmetic Mean of the it's inverse

Therefore,

$\begin{array}{rcl}\frac{{\displaystyle \frac{a+b+c}{b+c}}+{\displaystyle \frac{a+b+c}{c+a}}+{\displaystyle \frac{a+b+c}{a+b}}}{3}& \ge & {\left\{\frac{{\displaystyle \frac{b+c}{a+b+c}}+{\displaystyle \frac{c+a}{a+b+c}}+{\displaystyle \frac{a+b}{a+b+c}}}{3}\right\}}^{-1}\\ \frac{{\displaystyle \frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}}}{3}& \ge & {\left(\frac{2}{3}\right)}^{-1}\\ \frac{{\displaystyle \frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}}}{3}& \ge & \frac{3}{2}\\ \frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}& \ge & \frac{9}{2}\\ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+3& \ge & \frac{9}{2}\\ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}& \ge & \frac{9}{2}-3\\ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}& \ge & \frac{3}{2}\end{array}$

Hence, correct option is $\mathbf{\left(}\mathit{B}\mathbf{\right)}$