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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2020 · Mathematics · Sequences and Series

If a, b, c are positive numbers, then least value of (a+b+c)...

If a,b,ca, b, c are positive numbers, then least value of (a+b+c)(1a+1b+1c)is(a+b+c) \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)is

Options

  1. A1
  2. B6
  3. C9
  4. DNone

Correct answer

C. 9

Step-by-step solution

AM≥HM⇒a+b+c3≥31a+1b+1c⇒(a+b+c)(1a+1b+1c)≥9\mathrm{AM} \geq \mathrm{HM}\Rightarrow \frac{a+b+c}{3} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\Rightarrow(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 9
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