Question

If \[a,b,c\] are the numbers, then the least value of \[(a+b+c)\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right)\] is
A. \[1\]
B. \[6\]
C. \[9\]
D. None of these

Answer 1

Hint: In this question, we are to find the least value of the given expression. In the given expression, there are two series. They are of arithmetic and harmonic progressions. So, by using the relation between the arithmetic mean and the harmonic mean, we get the required least value.

Formula Used: If the series is an Arithmetic series, then
The $nth$ term of the series is ${{t}_{n}}=a+(n-1)d$
Where $a$ - First term
Here the common difference$d={{a}_{n}}-{{a}_{n-1}}$
The sum of $n$ terms of the series is
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$
Where ${{S}_{n}}$ - Sum of $n$ terms of the series

Complete step by step solution: Given numbers are \[a,b,c\].
The series formed by these numbers are
Arithmetic series: \[A=a+b+c\] and Harmonic series: $H=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$
Then, their means are
$\begin{align}
  & AM=\dfrac{a+b+c}{3} \\
 & HM=\dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} \\
\end{align}$
But we have
\[AM\ge GM\ge HM\]
Then, writing for the arithmetic and harmonic means, we get
\[\Rightarrow AM\ge HM\]
So, on substituting the mean we calculated above, we get
\[\begin{align}
  & \Rightarrow \dfrac{a+b+c}{3}\ge \dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} \\
 & \Rightarrow \left( a+b+c \right)\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right)\ge 9 \\
\end{align}\]
Since the given expression is greater than $9$, that will be the least value.

Option ‘C’ is correct

Note: Here we need to remember that, the means of the series should be compared but not the sums. By using the relation $AM\ge GM\ge HM$, we get the required value for the given expression. Since we got \[(a+b+c)\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right)\ge 9\], the value obtained will be the least value of the expression.