Daily

Random

Practice set

Baltic Way 1995 · Problem 8

Algebra

The real numbers a,ba, b and cc satisfy the inequalities ∣a∣≥∣b+c∣,∣b∣≥∣c+a∣|a| \geq|b+c|,|b| \geq|c+a| and ∣c∣≥∣a+b∣|c| \geq|a+b|. Prove that a+b+c=0a+b+c=0.

Change pool

When you’re ready

Review material becomes available with the next Daily.

Review

Topics

Equations and inequalities

Solutions

Solution

Solution:

Squaring both sides of the given inequalities we get

{a2≥(b+c)2b2≥(c+a)2c2≥(a+b)2\left\{ \begin{array}{l} a^{2} \geq (b+c)^{2} \\ b^{2} \geq (c+a)^{2} \\ c^{2} \geq (a+b)^{2} \end{array} \right.

Adding these three inequalities and rearranging, we get (a+b+c)2≤0(a+b+c)^{2} \leq 0. Clearly equality must hold, and we have a+b+c=0a+b+c=0.

Contest context

Results from Baltic Way 1995

9 teams

Mean score
4.4 / 5
Scores of 4 or 5
8 / 9
Estonia
5 / 5

Score distribution

01
10
20
30
40
58
All team scores
TeamScore
Poland5 / 5
Latvia5 / 5
Sweden5 / 5
Lithuania5 / 5
Denmark5 / 5
Finland5 / 5
St. Petersburg5 / 5
Estonia5 / 5
Iceland0 / 5