Question

We are giving the concept of A.M of $m^{th}$ power.Let $a,b>0$ and $a\ne b$ and let $m$ be a real number. Then
$\frac{a^m+b^m}{2}>{\left(\frac{a+b}{2}\right)}^m$ if $m\in R-[0,1]$
However, if $m\in (0,1)$ then $\frac{a^m+b^m}{2}<{\left(\frac{a+b}{2}\right)}^m$.
Obviously, if $m\in \{0,1\}$ then $\frac{a^m+b^m}{2}={\left(\frac{a+b}{2}\right)}^m$.
On the basis of the above information, answer the following questions:

If $a,b,c$ be positive real numbers, then the possible best option of values $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$ lie between:

• A. $(1,\infty )$
• B. $(0,\infty )$
• C. $[\frac{3}{2},\infty )$
• D. $[2,\infty )$

Answer 1

The correct option is C $[\frac{3}{2},\infty )$

$\because \frac{{\left(\frac{b+c}{a+b+c}\right)}^{-1}+{\left(\frac{c+a}{a+b+c}\right)}^{-1}+{\left(\frac{a+b}{a+b+c}\right)}^{-1}}{3}\ge {\left(\frac{\frac{b+c}{a+b+c}+\frac{c+a}{a+b+c}+\frac{a+b}{a+b+c}}{3}\right)}^{-1}$

$={\left(\frac{2(a+b+c)}{3(a+b+c)}\right)}^{-1}=\frac{3}{2}$

$\Rightarrow {\left(\frac{b+c}{a+b+c}\right)}^{-1}+{\left(\frac{c+a}{a+b+c}\right)}^{-1}+{\left(\frac{a+b}{a+b+c}\right)}^{-1}\ge \frac{9}{2}$

$\Rightarrow \frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\ge \frac{9}{2}$

$\Rightarrow \frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\ge \frac{9}{2}$

$\Rightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{9}{2}-3\ge \frac{9-6}{2}=\frac{3}{2}$

$\therefore \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$

$\therefore \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\in [\frac{3}{2},\infty )$