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# If a,b,c,d are positive real number , then least value of $\;(a+b+c+d)\;(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})\;$ is :

$(a)\;4\qquad(b)\;8\qquad(c)\;16\qquad(d)\;2$

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A)
Answer : (c) 16
Explanation : $\;AM\;\geq\;GM$
$\large\frac{a+b+c+d}{4}\;\geq\;(abcd)^{\frac{1}{4}}$
$and\;\frac{1}{4}\;(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})\;\geq\;(\large\frac{1}{abcd})^{\frac{1}{4}}$
Multiplying both
$\frac{1}{16}\;(a+b+c+d)\;(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})\;\geq\;1$
Least value of
$\;(a+b+c+d)\;(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})=\frac{1}{16}.$