## Sunday, April 2, 2006

### Absolutely! Topic: Algebra/Inequalities. Level: AIME/Olympiad.

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.

--------------------

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.

--------------------

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 $0 < p<15$.