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How to check that the arithmetic mean of any subset $S\ne\{1\}$ of $A=\{1,2,...,n\}$ is at least $\frac 32$?

I don't have any idea where to start this. It is clear that this is right and I need to demonstrate because it would help me solve another problem.

I think is something using the inequality of means or something.

Hope one of you can help me. Thank you!

Adam Rubinson
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IONELA BUCIU
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    Consider a subset $S$. Suppose its mean is $x$. Then $S \cup {y}$ has a mean greater than $x$ iff $y>x$. Otherwise new mean lies between $x$ and $y$. Now, what is the smallest possible mean for any $S$? – D S Dec 20 '23 at 17:51
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    Hint: if $S$ has $k$ elements, their mean is at least $\frac1k\sum_{j=1}^kj$. By the way, you should also require $S\ne\emptyset$ so the mean exists. – J.G. Dec 20 '23 at 17:58

3 Answers3

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Let $\{x_1,\ldots,x_k\}$ be a subset of $A$ with cardinal $k\geq 2$. Then, $$\frac{x_1+\ldots+x_k}{k}\geq\frac{1+\ldots+k}{k}=\frac{k(k+1)/2}{k}=(k+1)/2\geq(2+1)/2=3/2.$$

Deif
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  • Thank you so much for your idea, but how did you get to that inequality? – IONELA BUCIU Dec 20 '23 at 18:17
  • ${1,2,\ldots,k}$ is the subset of $A$ with $k$ elements with minimum sum. Then any subset ${x_1,\ldots,x_k}$ of $A$ with $k$ elements has its sum greater or equal to the sum of ${1,2,\ldots,k}$. Feel free to ask if you have any questions. – Deif Dec 20 '23 at 18:46
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    Now I understand! Thank you so much! – IONELA BUCIU Dec 20 '23 at 19:01
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You don't really need to use anything fancy. The mean of your set is going to be bigger than the mean you'd have if you replaced all the numbers that aren't 1 with 2, right?

Now, what's the mean of a single 1 and a bunch of 2s?

(I guess you're using the fact that if two sets have the same number of elements, and you can pair them up so that the elements from the first set are always at least as big as the corresponding elements from the second set, then the sum (and thus the average) of the first set is at least as big as the second set's. Is that what you intend by "the inequality of means"? I'm not familiar with that name.)

JonathanZ
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Let $k:= \vert S \vert\ (\leq n),\ $ and write $\ S=\lbrace{a_1, a_2, \ldots, a_k \rbrace},\ $ where $\ a_1 < a_2 < \ldots < a_k.$

If $a_1 \geq 2,$ then $\displaystyle\sum_{ i=1 }^{k} a_i \geq 2k,\ \implies \frac{\displaystyle\sum_{ i=1 }^{k} a_i }{k} \geq \frac{2k}{k}=2\geq \frac{3}{2}.$

Else, $a_1 = 1,$ and since we must avoid $S=\lbrace{1\rbrace},\ $ we must have $k\geq 2$ and $a_2\geq 2,\ $ in which case $\displaystyle\sum_{ i=1 }^{k} a_i = a_1 + \sum_{ i=2 }^{k} a_i = 1 + \sum_{ i=2 }^{k} a_i \geq 1 + 2(k-1) = 2k-1,\ \implies \frac{\displaystyle\sum_{ i=1 }^{k} a_i }{k} \geq \frac{2k-1}{k} = 2-\frac{1}{k}\geq \frac{3}{2},\ $ because $k\geq 2.$

Adam Rubinson
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