Proving average is greater

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 $$n$$.

• What have you tried? Where are you stuck? Sep 26 '19 at 3:10
• I tried induction unsuccessfully Sep 26 '19 at 3:17
• You don't need induction, just the definition of average. Sep 26 '19 at 3:22
• If they are all greater than $n$ then the smallest is larger than $n$. And the smallest can't be greater than the average. Sep 26 '19 at 3:34

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

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

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