Question

If $\frac{1}{P_1}+\frac{1}{P_2}+\frac{1}{P_3}=\frac{1}{4}$, then the least value of $P_1{P}_2{P}_3$ is?

• A. $1728$
• B. $216$
• C. $144$
• D. $\frac{1}{12}$

Answer 1

The correct option is A $1728$

Applying $A.M\ge G.M$ on $\frac{1}{P_1},\frac{1}{P_2},\frac{1}{P_3}$
We get
$\frac{\frac{1}{P_1}+\frac{1}{P_2}+\frac{1}{P_3}}{3}\ge {(\frac{1}{P_1}.\frac{1}{P_2}.\frac{1}{P_3})}^{\frac{1}{3}}\Rightarrow \frac{1}{12}\ge {\left(\frac{1}{P_1.P_2.P_3}\right)}^{\frac{1}{3}}\Rightarrow {\left(\frac{1}{12}\right)}^3\ge \frac{1}{P_1.P_2.P_3}\Rightarrow P_1.P_2.P_3\ge 1728$