**Problem**: (1999 USAMO - #4) Let $a_{1}, a_{2}, \ldots, a_{n}$ ($n > 3$) be real numbers such that

$a_{1} + a_{2} + \cdots + a_{n} \geq n \qquad$ and $\qquad a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2} \geq n^{2}$.

Prove that $\max(a_{1}, a_{2}, \ldots, a_{n}) \geq 2$.

**Solution**: To simply things, let $ x_i = 2-a_i $. We wish to show that there exists an $ x_i \le 0 $.

Our conditions become

$ (2-x_1)+(2-x_2)+\cdots+(2-x_n) \ge n \Rightarrow x_1+x_2+\cdots+x_n \le n $.

$ (2-x_1)^2+(2-x_2^2)+\cdots+(2-x_n)^2 \ge n^2 $.

Assume for the sake of contradiction that $ x_i > 0 $ for all $ i $ and let $ x_1+x_2+\cdots+x_n = k \le n $. We have

$ (2-x_1)^2+(2-x_2)^2+\cdots+(2-x_n^2) = (x_1^2+x_2^2+\cdots+x_n^2) - 4(x_1+x_2+\cdots+x_n)+4n \ge n^2 $, or

$ (x_1^2+x_2^2+\cdots+x_n^2) - 4(x_1+x_2+\cdots+x_n) \ge n^2-4n $.

But we have $ x_1^2+x_2^2+\cdots+x_n^2 < (x_1+x_2+\cdots+x_n)^2 = k^2 $, so

$ (x_1^2+x_2^2+\cdots+x_n^2) - 4(x_1+x_2+\cdots+x_n) < k^2-4k $.

However, since $ n \ge k $ and $ n \ge 4 $, we have $ n^2-4n \ge k^2-4k $ (looking at the parabola it is easy to see; $ x^2-4x $ has zeros at $ x = 0, 4 $). Therefore

$ (x_1^2+x_2^2+\cdots+x_n^2) - 4(x_1+x_2+\cdots+x_n) < k^2-4k \le n^2-4n $.

But we also had

$ (x_1^2+x_2^2+\cdots+x_n^2) - 4(x_1+x_2+\cdots+x_n) \ge n^2-4n $,

so that gives us a contradiction. Hence our assumption must be false and there exists an $ x_i \le 0 \Rightarrow a_i \ge 2 $ as desired. QED.

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Comment: This wasn't a particularly difficult inequality, but it has some key ideas. Using all parts of the question is important (in this case $ n > 3 $ is actually relevant). Another note is that this didn't really require any of the classical inequalities, just algebraic manipulation. Lastly, the crucial step was setting converting the real $ a_i $ to positive $ x_i $ which makes things a whole lot nicer.

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Practice Problem: Go take the 2006 Mock AIME 5 below.

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