Question

If $a>0,b>0$and$a+b=1$. Then $a^{\frac{1}{3}}+b^{\frac{1}{3}}$ is

• A. $>2^{\frac{2}{3}}$
• B. $<2^{\frac{2}{3}}$
• C. $>4^{\frac{2}{3}}$
• D. $<4^{\frac{2}{3}}$

Answer 1

The correct option is B

$<2^{\frac{2}{3}}$

Explanation for the correct option:

Finding value of greater or less than,

Since A.M of ${\left(\frac{1}{3}\right)}^{th}$ powers $<{\left(\frac{1}{3}\right)}^{th}$ power of A.M

$$\begin{gathered} \left[\frac{(a^{{\displaystyle \frac{1}{3}}}+b^{{\displaystyle \frac{1}{3}}})}{2}\right]<{\left[\frac{(a+b)}{2}\right]}^{\frac{1}{3}} \\ [\frac{(a^m+b^m)}{2}<{\left[\frac{(a+b)}{2}\right]}^m,if0<m<1] \\ {a}^{\frac{1}{3}}+b^{\frac{1}{3}}<2{\left(\frac{1}{2}\right)}^{\frac{1}{3}} \\ {a}^{\frac{1}{3}}+b^{\frac{1}{3}}<2^{\frac{2}{3}} \end{gathered}$$

Hence, the correct option is $\left(B\right)$.