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Sum of fifty positive nos. is 1.

Find maximum value of sum of their inverse.

I have no idea how to solve this question... do not mark it as off topic or anything...

Maybe we should use AM>=GM?

geek101
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3 Answers3

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Let $x_1,x_2,\ldots,x_{50}$ positive numbers such that their sum is $1$.

We can prove that the sum $\frac{1}{x_1}+\frac{1}{x_2}+\ldots+\frac{1}{x_{50}}$ is not bounded.

Let $M>1$, then by putting $x_1=\frac{1}{2M}$ and $x_2=x_3=\dots =x_{50}=\frac{1}{49}\left(1-\frac{1}{2M}\right)$ we get $$\frac{1}{x_1}+\frac{1}{x_2}+\ldots+\frac{1}{x_{50}}=2M+\underbrace{\frac{98}{2M-1}+\frac{98}{2M-1}+\ldots+\frac{98}{2M-1}}_{49\text{ times}}>M$$

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The maximum is infinity. Take one of the number to approach $0$ and all the others sum to approach $1$.

Hans
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Assume that the numbers are equal. Therefore the number must be $1/50$ When you say inverse if you mean reciprocal then the maximum value of the reciprocal becomes $50*50=2500$ Science I assumed the numbers are equal their sums are the maximum

Socre
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  • shouldnt minimum be 2500? – geek101 Mar 28 '15 at 18:30
  • Sorry, what was I thinking, we can set every 49 numbers to zero by letting the denominator to aproach to infinity and the last number to approach to one. The reciprocal being $\infinity$. 2500 being the minimum value because 50 being the maximum denominator for every function.... Sorry, – Socre Mar 28 '15 at 18:38
  • no problem.... :) – geek101 Mar 28 '15 at 18:48