Question

If we have the values $a > 0$, $b > 0$ and $a + b = 1$. Then ${a^{\dfrac{1}{3}}} + {b^{\dfrac{1}{3}}}$ is
(A) $> {2^{\dfrac{2}{3}}}$
(B) $< {2^{\dfrac{2}{3}}}$
(C) $> {4^{\dfrac{2}{3}}}$
(D) $< {4^{\dfrac{2}{3}}}$

Answer 1

Here, we will first use one of the properties of arithmetic mean that is
$\dfrac{{\left( {{a^n} + {b^n}} \right)}}{2} < {\left( {\dfrac{{a + b}}{2}} \right)^n}{\text{if 0 < n < 1}}$
Then in this equation we will put the required value of $n = \dfrac{1}{3}$. Also, it is given in the question that $a + b = 1$, so we will put this value in the right-hand side of the above equation to find the required value.

Complete step-by-step solution:
By the properties of arithmetic mean we can write:
$\left( {\dfrac{{{a^n} + {b^n}}}{2}} \right) < {\left( {\dfrac{{a + b}}{2}} \right)^n} - - - (1)$ $0 < n < 1$
To check whether this property is true or not we can use mathematical induction or we can even check by just putting random values of the variables. Let us take some random values of $a,b,n$.
Let,
$a = 8$
$b = 27$
$n = \dfrac{1}{3}$
So, the left-hand side of the above equation becomes
$\left( {\dfrac{{{a^n} + {b^n}}}{2}} \right) = \left( {\dfrac{{{8^{\dfrac{1}{3}}} + {{27}^{\dfrac{1}{3}}}}}{2}} \right)$
$\Rightarrow \left( {\dfrac{{{a^n} + {b^n}}}{2}} \right) = \left( {\dfrac{{2 + 3}}{2}} \right)$
On further simplification it becomes
$\Rightarrow \left( {\dfrac{{{a^n} + {b^n}}}{2}} \right) = 2.5 - - - (2)$
Similarly, on putting the values in right-hand side of the equation, we get
${\left( {\dfrac{{a + b}}{2}} \right)^n} = {\left( {\dfrac{{8 + 27}}{2}} \right)^{\dfrac{1}{3}}}$
$\Rightarrow {\left( {\dfrac{{a + b}}{2}} \right)^n} = {\left( {17.5} \right)^{\dfrac{1}{3}}}$
By using the calculator, we will find the value of right-hand side of the above equation
$\Rightarrow {\left( {\dfrac{{a + b}}{2}} \right)^n} = 2.5962 - - - (3)$
So, from equation $(2)$ and equation $(3)$ it is clear that equation $(1)$ is true.
Since, in the question we are asked to find the value of ${a^{\dfrac{1}{3}}} + {b^{\dfrac{1}{3}}}$ so we will take $n = \dfrac{1}{3}$ in equation $(1)$.
So, equation $(1)$ becomes,
$\left( {\dfrac{{{a^{\dfrac{1}{3}}} + {b^{\dfrac{1}{3}}}}}{2}} \right) < {\left( {\dfrac{{a + b}}{2}} \right)^{\dfrac{1}{3}}}$
Now in the right-hand side of the above equation we will put the value of $a + b = 1$, which is given in the question. Now the equation becomes;
$\Rightarrow \left( {\dfrac{{{a^{\dfrac{1}{3}}} + {b^{\dfrac{1}{3}}}}}{2}} \right) < {\left( {\dfrac{1}{2}} \right)^{\dfrac{1}{3}}}$
On shifting $2$ to the right-hand side, we get
$\Rightarrow \left( {{a^{\dfrac{1}{3}}} + {b^{\dfrac{1}{3}}}} \right) < 2{\left( {\dfrac{1}{2}} \right)^{\dfrac{1}{3}}}$
On further simplification the above equation becomes
$\Rightarrow \left( {{a^{\dfrac{1}{3}}} + {b^{\dfrac{1}{3}}}} \right) < {2^{\dfrac{2}{3}}}$
Therefore, ${a^{\dfrac{1}{3}}} + {b^{\dfrac{1}{3}}}$ is less than ${2^{\dfrac{2}{3}}}$ i.e., $\left( {{a^{\dfrac{1}{3}}} + {b^{\dfrac{1}{3}}}} \right) < {2^{\dfrac{2}{3}}}$.
Hence, option (B) is correct.

Note: Here, one important point to note is that for equation $(1)$ to be true, the value of $n$ should be between zero and one i.e., $0 < n < 1$. Also note that for $n = 0{\text{ and }}n = 1$, that is for the upper and lower limits of $n$, equality holds means that the left-hand side of equation $(1)$ becomes equal to the right-hand side.