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

Question

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!

Answer 1

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

Answer 2

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.)

Answer 3

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 $k\geq 2.$