**Problem**: (Math Olympiad Treasures - 1.3.13) Suppose the sequence $ a_1, a_2, \ldots, a_n $ satisfies the following conditions:

$ a_1 = 0 $,$ |a_2| = |a_1+1| $, $ \ldots $, $ |a_n| = |a_{n-1}+1| $.

Prove that

$ \frac{a_1+a_2+\cdots+a_n}{n} \ge -\frac{1}{2} $.

**Solution**: Consider squaring the relations given in the sequence to obtain

$ a_2^2 = (a_1+1)^2 = a_1^2+2a_1+1 $

$ a_3^2 = (a_2+1)^2 = a_2^2+2a_2+1 $

$ \ldots $

$ a_{n+1}^2 = (a_n+1)^2 = a_n^2+2a_n+1 $.

Adding them up, we have

$ a_2^2+a_3^2+\cdots+a_{n+1}^2 = a_1^2+a_2^2+\cdots+a_n^2+2(a_1+a_2+\cdots+a_n)+n $, or

$ a_{n+1}^2 = 2(a_1+a_2+\cdots+a_n)+n \ge 0 $.

Then we have

$ \frac{a_1+a_2+\cdots+a_n}{n} \ge -\frac{1}{2} $,

as desired. QED.

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Comment: Absolute values are easy to get rid of by squaring things, usually giving us nice results. This method can be used in a variety of problems, but many times with sequences recursively defined we see it at its best. The following AIME problem can be solved this way.

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Practice Problem: (2006 AIME I - #15) Given that a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge 1$, find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|$.

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