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If $a, b, c, d$ and $e$ are positive real numbers, then the minimum value of $(a+b+c+d+e)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e})$ is

  1. $25$
  2. $5$
  3. $125$
  4. Cannot be determined.
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3 Answers

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Its answer is A) 25, when all a,b,c,d and e are=1, in all other cases It will be greater than that. Try Hit and trial and option elimination
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1. 25

Because the if we increase the value of a,b,c,d,e then value of part-1 of equation increases

and if we decrease the value the part-2 will be increases so this function is minimum at a=b=c=d=e=1;

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