Question

For three distinct positive numbers $p,q\mathrm{and}r$, if $p+q+r=a$, then:

• A. $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}>\frac{18}{a}$
• B. $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}>\frac{9}{a}$
• C. $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}<\frac{3}{a}$
• D. $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}<\frac{6}{a}$

Answer 1

The correct option is B

$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}>\frac{9}{a}$

Explanation for the correct option:

Finding the value of $\frac{\mathbf{1}}{\mathbf{p}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{q}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{r}}$:

As per the $AM-GM$ inequality property, the arithmetic mean of three distinct positive numbers $>$ their geometric mean.

$$\begin{gathered} \Rightarrow \frac{p+q+r}{3}>{\left(p\times q\times r\right)}^{1/3}\left[\mathrm{by}AM=\frac{\Sigma n}{n}andGM=\sqrt[n]{\mathrm{Product}\mathrm{of}n\mathrm{numbers}}\right] \\ \Rightarrow p+q+r>3{\left(pqr\right)}^{1/3}...\left(1\right) \end{gathered}$$

Similarly,

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}+{\displaystyle \frac{1}{r}}}{3}>{\left(\frac{1}{p}\times \frac{1}{q}\times \frac{1}{r}\right)}^{1/3} \\ \Rightarrow \frac{1}{p}+\frac{1}{q}+\frac{1}{r}>3{\left(\frac{1}{pqr}\right)}^{1/3}...\left(2\right) \end{gathered}$$

By multiplying $\left(1\right)$ and $\left(2\right)$, we get

$$\begin{gathered} \left(p+q+r\right)\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\right)>3{\left(pqr\right)}^{1/3}\times 3{\left(\frac{1}{pqr}\right)}^{1/3} \\ \Rightarrow \left(p+q+r\right)\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\right)>9{\left(1\right)}^{1/3}\left[\mathrm{by}{a}^m\times b^m={\left(ab\right)}^m\right] \\ \Rightarrow a\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\right)>9(\mathrm{given}p+q+r=a) \\ \Rightarrow \frac{1}{p}+\frac{1}{q}+\frac{1}{r}>\frac{9}{a} \end{gathered}$$

Therefore, the correct answer is option B.