Proof Verification - $\exists a \in S (a\ge S_a)$ I wanted to prove that $\exists a \in S (a\ge S_a)$ where $S$ is an finite set of real numbers with order $n$ and $S_a$ is the average of the set. This is my proof so far:
Assume $a_i = a_k, i,k \text{ are integers and } \in [1,n]$. Also assume $a_i < S_a = \frac{\sum a_i}{n}$
The last statement implies $na_i < \sum a_i = na_i$ which is a contradiction.
Now, can we generalize this statement for $a_i \neq a_k$ by saying that the amount $S_a - a_i$ for some $a_i<S_a$ is compensated by adding it to some $a_k > S_a$?

Let  $$S=\{a_{1},a_{2},........a_{n}\}$$  be  the  finite  set.      Assume  there  is  no  element  $a$   in  $S$  such  that  $a\ge S_{a}$. Now $$\sum_{i=1}^{n} a_{i} = n.S_{a}$$ . But  according  to  our  assumption each  $a_{i} \lt S_{a}$.  Hence $$a_{1}+a_{2}+.....a_{n} \lt {S_{a}+S_{a}+......S_{a}}_{\{\ \ added\ \   n \ \  times}\} $$ i.e. $$\sum_{i=1}^{n} a_{i} \lt n.S_{a}$$   
Thus  the  contradiction in  general  case.  So  there  must  exist  an  $a_{m}$  for  some  $m\le n$  such that  $$a_{m} \ge S_{a}$$
