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 $a$ and $b$ are positive $(a\ne b)$ and $a+b=1$ and if $A={(a+\frac{1}{a})}^2+{(b+\frac{1}{b})}^2,$ then the correct statement is:

• A. $A>8$
• B. $A<8$
• C. $A>\frac{25}{2}$
• D. $A<\frac{25}{2}$

Answer 1

Answer: (C)

$A>\frac{25}{2}$

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

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

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

As $a+b=1$ we get

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

Again, $\frac{a^{-1}+b^{-1}}{2}>{\left(\frac{a+b}{2}\right)}^{-1}={\left(\frac{1}{2}\right)}^{-1}=2$

$\therefore a^{-1}+b^{-1}>4$
or $1+a^{-1}+b^{-1}>1+4=5$
Squaring both sides, we get

${(1+a^{-1}+b^{-1})}^2>5^2=25$
or $\frac{1}{2}{(1+a^{-1}+b^{-1})}^2>\frac{25}{2}$ .......$\left(2\right)$

From eqns$\left(1\right)$ and $\left(2\right)$, we get

${(a+\frac{1}{a})}^2+{(b+\frac{1}{b})}^2>\frac{25}{2}$
or $A>\frac{25}{2}$