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Question

If $$\displaystyle \frac{1}{P_{1}}+\frac{1}{P_{2}}+\frac{1}{P_{3}}=\frac{1}{4}$$, then the least value of $$\displaystyle P_{1} P_{2} P_{3}$$ is?


A
1728
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B
216
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C
144
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D
112
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Solution

The correct option is A $$1728$$
Applying $$A.M\ge G.M$$ on $$\cfrac { 1 }{ { P }_{ 1 } } ,\cfrac { 1 }{ { P }_{ 2 } } ,\cfrac { 1 }{ { P }_{ 3 } } $$
We get
$$\cfrac { \cfrac { 1 }{ { P }_{ 1 } } +\cfrac { 1 }{ { P }_{ 2 } } +\cfrac { 1 }{ { P }_{ 3 } }  }{ 3 } \ge { \left( \cfrac { 1 }{ { P }_{ 1 } } .\cfrac { 1 }{ { P }_{ 2 } } .\cfrac { 1 }{ { P }_{ 3 } }  \right)  }^{ \cfrac { 1 }{ 3 }  }\\ \Rightarrow \cfrac { 1 }{ 12 } \ge { \left( \cfrac { 1 }{ { P }_{ 1 }.{ P }_{ 2 }.{ P }_{ 3 } }  \right)  }^{ \cfrac { 1 }{ 3 }  }\\ \Rightarrow { \left( \cfrac { 1 }{ 12 }  \right)  }^{ 3 }\ge \cfrac { 1 }{ { P }_{ 1 }.{ P }_{ 2 }.{ P }_{ 3 } } \\ \Rightarrow { P }_{ 1 }.{ P }_{ 2 }.{ P }_{ 3 }\ge 1728$$

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