Math Help - How to disprove that all numbers A1,A2,A3, A4....An are less than their average?

1. How to disprove that all numbers A1,A2,A3, A4....An are less than their average?

Its something very simple yet I cant figure it out. I know that if all a are smaller than their average, then you can show that this average is smaller than itself. But how to actually do that? I'm new to Discrete Mathematics and am having a hard time so please help me out. Thanks

2. Re: How to disprove that all numbers A1,A2,A3, A4....An are less than their average?

Suppose

\begin{align*}a_1&<\frac{a_1+\dots+a_n}{n}\\ &\hspace{1cm}\cdots \\a_n&<\frac{a_1+\dots+a_n}{n}\end{align*}

3. Re: How to disprove that all numbers A1,A2,A3, A4....An are less than their average?

X (average) = (A1+A2+...An)/n

A1+A2+...An < X+X+...Xn (upto n times)

Divide both sides by n:
A1/n +A2/n... An/n < X/n +X/n ... Xn/n

Am I doing it right? What next? Take 1/n common on both sides? I would still have many X+X+X...Xn left on the right side wont I?? Or I'm completely missing something?

4. Re: How to disprove that all numbers A1,A2,A3, A4....An are less than their average?

Originally Posted by XingPing
X (average) = (A1+A2+...An)/n

A1+A2+...An < X+X+...Xn (upto n times)

Divide both sides by n:
A1/n +A2/n... An/n < X/n +X/n ... Xn/n
You are done now because the left-hand side is (A1 + ... + An)/n = X, and the right-hand side is X/n + ... + X/n (n times) = X, so the last line says that X < X, a contradiction.

Two remarks. First, I assume by X+X+...Xn you mean X + ... + X (n times). It is not clear what Xn is: a product of X and n (which is legal, but does not fit the context) or X with a subscript n (which does not make sense).

Second, in the beginning you need to say, "Assume, towards contradiction, that A1 < X, ...., An < X. Then...". Otherwise, it is not clear from where A1 + A2 + ... + An < X + X + ... + X follows.

It is also not clear to me why you ignored the hint in post #2.