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 $a,b$ be positive and $a+b=1(a\ne b)$ and if $A=\sqrt[3]{3a}+\sqrt[3]{3b}$ then the correct statement is:

• A. $A>2^{\frac{2}{3}}$
• B. $A=\frac{2^{\frac{2}{3}}}{3}$
• C. $A<2^{\frac{2}{3}}$
• D. $A=2^{\frac{2}{3}}$

Answer 1

The correct option is C $A<2^{\frac{2}{3}}$

Take $m=\frac{1}{3}\in (0,1)$

Then $\frac{a^{\frac{1}{3}}+b^{\frac{1}{3}}}{2}<{\left(\frac{a+b}{2}\right)}^{\frac{1}{3}}={\left(\frac{1}{2}\right)}^{\frac{1}{3}}=\frac{1}{2^{\frac{1}{3}}}$ ..... Since $(a+b=1)$

$\therefore \frac{a^{\frac{1}{3}}+b^{\frac{1}{3}}}{2}<\frac{1}{2^{\frac{1}{3}}}$

$\Rightarrow A<\frac{2}{2^{\frac{1}{3}}}=2^{1-\frac{1}{3}}=2^{\frac{2}{3}}$ ...... $(\text{Using the law of exponents}\frac{a^m}{a^n}=a^{m-n})$

Thus, $A<2^{\frac{2}{3}}$