Question

Let $a_1,a_2...a_n$ be a set of $n$ positive real number (not all equal),
Let $m$be a real number then the arithmetic mean of $m^{th}$ power is counted as
(i):$\frac{a_1^m+a_2^m+a_3^m+...+a_1^m}{n}>{\left(\frac{a_1+a_2+...+a_n}{n}\right)}^m$ if the real number $m\in R-[0,1]$
(ii): $\frac{a_1^m+a_2^m+a_3^m+...+a_1^m}{n}<{\left(\frac{a_1+a_2+...+a_n}{n}\right)}^m$ if $m\in (0,1)$ and equality holds if $m=0$ or $m=1$.

On the basis of above information answer the following questions,Let $a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}+d^{\frac{2}{3}}+e^{\frac{2}{3}}+f^{\frac{2}{3}}+g^{\frac{2}{3}}+h^{\frac{2}{3}}=4$ and $a+b+c+d+e+f+g+h=\lambda$ then

• A. $\lambda <2^{\frac{3}{2}}$
• B. $\lambda <2^{\frac{1}{3}}$
• C. $\lambda >2^{\frac{3}{2}}$
• D. None of these

Answer 1

The correct option is C $\lambda >2^{\frac{3}{2}}$

In the problem $m=\frac{2}{3},n=8$
$\therefore$ Using $A.M$ of $m^{th}$ powers $<$ $m^{th}$ powers of $A.M$
$\Rightarrow \frac{a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}+d^{\frac{2}{3}}+e^{\frac{2}{3}}+f^{\frac{2}{3}}+g^{\frac{2}{3}}+h^{\frac{2}{3}}}{8}<{\left(\frac{(a+b+c+d+e+f+g+h)}{8}\right)}^{\frac{2}{3}}$
$\Rightarrow \frac{1}{2}<{\left(\frac{\lambda }{8}\right)}^{\frac{2}{3}}\Rightarrow {\left(\frac{1}{2}\right)}^{\frac{3}{2}}<\frac{\lambda }{8}$
$\therefore \lambda >{8.8}^{-1.2}$
$\Rightarrow \lambda >8^{\frac{1}{2}}=2^{\frac{3}{2}}$
$\therefore \lambda >2^{\frac{3}{2}}$

Ans: C