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If all elements in the set of positive real numbers are all greater than a arbitrary positive number, prove that the average of that set is greater than said arbitrary positive number. In other words if a set $S$ consisted of positive numbers and all elements of $S$ were greater than a real positive number $n$, then prove that the average of $S$ is greater than $n$.

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    $\begingroup$ What have you tried? Where are you stuck? $\endgroup$
    – angryavian
    Sep 26 '19 at 3:10
  • $\begingroup$ I tried induction unsuccessfully $\endgroup$ Sep 26 '19 at 3:17
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    $\begingroup$ You don't need induction, just the definition of average. $\endgroup$
    – saulspatz
    Sep 26 '19 at 3:22
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    $\begingroup$ If they are all greater than $n$ then the smallest is larger than $n$. And the smallest can't be greater than the average. $\endgroup$
    – fleablood
    Sep 26 '19 at 3:34
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Hint: Pick any $x_k \in S$ and remember $y < x_k$

Now add all the items in $S$ together to form the inequality

$$ \sum_{i = 0} ^ n y < \sum_{i=0}^n x_i \implies ny < \sum_{i=0}^n x_i $$

Hopefully you can see where to go from there

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You've heard the joke about "all the children are above average", haven't you? Some of the elements must be at most average and if $n$ is less than that element, $n$ is below average.

1) $n < \min(S) \le avg(S)$

What more needs to be said?

Well, I suppose we should prove it's not possible for all elements of a set to be above average.

If $S$ has $m$ elements and they are $a_1 \le a_2 \le a_3 .... \le a_m$ and if $n < a_1$ then

$m \times n < m\times a_1 = a_1 + a_1 + a_1 + ..... + a_1\le a_1 + a_2 + a_3 + ... + a_m$

So

$\frac {m\times n}m < \frac {m\times a_1}m \le \frac{a_1+...+a_m}m$

And so

$n < a_1 \le avg(S)$.

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