I'm looking for verification of my proof.
Suppose $\exists \ i,j \ $ such that $s_i \neq s_j$. Prove $\exists \ k \ $ such that $s_k < A$
Here $A = \frac{s_1+s_2+ ... + s_n}{n}$ is the average of the sequence.
Proof:
Assume, for contradiction, $\neg \ \exists \ k (s_k < A)$. That is, assume $\forall \ k(s_k \geq A)$. But, since the hypothesis says that not all members of the sequence are equal, then we can't have $\forall \ k(s_k = A)$. So we must only consider the case where $\forall \ k(s_k > A)$.
By assumption:
$s_1 > A$
$s_2 > A$
...
$s_n > A$
Adding these inequalities, we get: $s_1+s_2+...+s_n > nA$. Dividing by n, we get $\frac{s_1+s_2+ ... + s_n}{n} > A$. This contradicts the definition of A. Therefore, we've proven by contradiction that $\exists \ k \ $ such that $s_k < A$. $\square$