**Problem**: Let $ a,b,c $ be reals such that $ |a| \ge |b+c| $, $ |b| \ge |c+a| $, and $ |c| \ge |a+b| $. Prove that $ a+b+c = 0 $.

**Solution**: WLOG, we may assume $ |a| \le |b| \le |c| $ because the problem is symmetric. Then $ |b+c| \le |a| \le |b| $ so $ b $ and $ c $ have opposite signs (either can be zero). Also $ |c+a| \le |b| \le |c| $ so $ a $ and $ c $ have opposite signs also (or zero). WLOG, assume $ a,b \le 0 $ and $ c \ge 0 $.

Now we proceed by contradiction. Suppose that $ a+b+c > 0 $ or $ a+b+c < 0 $.

CASE 1: $ a+b+c > 0 $

Then $ a+c > -b $ and both sides are positive so $ |a+c| > |b| $, contradiction.

CASE 2: $ a+b+c < 0 $

Then $ a+b < -c $ and both sides are negative so $ |a+b| > |c| $, contradiction.

Hence our assumption is false and we must have $ a+b+c = 0 $, as desired. QED.

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Comment: The idea was to exploit the symmetry of the problem effectively to bring about a contradiction. The algebra all works out nicely, too. For an alternative solution that preserves symmetry, try squaring the inequalities.

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Practice Problem: (1983 AIME - #2) Let $f(x)=|x-p|+|x-15|+|x-p-15|$, where $p \leq x \leq 15$. Determine the minimum value taken by $f(x)$ by $x$ in the interval $0 < p<15$.

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