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,If $a$ & $b$ are positive real numbers and $a+b=128$, then maximum of $a^{1/3}+b^{1/3}$ is less than

• A. 6
• B. 8
• C. 10
• D. 12

Answer 1

The correct option is B 8

Using $A.M.$ of $m^{th}$ powers $<m^{th}$ powers of $A.M$ for $m=\frac{1}{3}$

$\frac{a^{1/3}+b^{1/3}}{2}<{\left(\frac{a+b}{2}\right)}^{1/3}\Rightarrow \frac{a^{1/3}+b^{1/3}}{2}<{\left(\frac{128}{2}\right)}^{1/3}$

$a^{1/3}+b^{1/3}<8$

Hence choice (B) is correct answer.