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

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

Suppose

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

Add all these inequalities.

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?

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

Quote:

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.