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 $a,b,c$ are positive real numbers but not all equal such that $a+b+c=1,$ then best option of values $\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}$ lie between:

• A. $(\frac{3}{2},\infty )$
• B. $(1,\infty )$
• C. $(0,\infty )$
• D. None of the above

Answer 1

The correct option is B $(1,\infty )$

Using the concept of A.M of $m^{th}$ power

$\frac{a^m+b^m}{2}>{\left(\frac{a+b}{2}\right)}^m$ we have

$\frac{a^2+b^2}{2}>{\left(\frac{a+b}{2}\right)}^2$

$\Rightarrow \frac{a^2+b^2}{2}>\frac{a+b}{2}\times \frac{a+b}{2}$

$\Rightarrow \frac{a^2+b^2}{a+b}>\frac{a+b}{2}$ ........$\left(1\right)$

$\frac{b^2+c^2}{2}>{\left(\frac{b+c}{2}\right)}^2$

$\Rightarrow \frac{b^2+c^2}{2}>\frac{b+c}{2}\times \frac{b+c}{2}$

$\Rightarrow \frac{b^2+c^2}{b+c}>\frac{b+c}{2}$ ........$\left(2\right)$

and $\frac{c^2+a^2}{2}>{\left(\frac{c+a}{2}\right)}^2$

$\Rightarrow \frac{c^2+a^2}{2}>\frac{c+a}{2}\times \frac{c+a}{2}$

$\Rightarrow \frac{c^2+a^2}{c+a}>\frac{c+a}{2}$ ........$\left(3\right)$

Adding equations $\left(1\right),\left(2\right),\left(3\right)$ we get

$\frac{b^2+c^2}{b+c}+\frac{a^2+b^2}{a+b}+\frac{c^2+a^2}{c+a}>\frac{b+c}{2}+\frac{a+b}{2}+\frac{c+a}{2}>a+b+c$

As $a+b+c=1$ we have

$\frac{b^2+c^2}{b+c}+\frac{a^2+b^2}{a+b}+\frac{c^2+a^2}{c+a}>1$

or $\frac{b^2+c^2}{b+c}+\frac{a^2+b^2}{a+b}+\frac{c^2+a^2}{c+a}\in (1,\infty )$