Question

If a, b, c are three real numbers such that | a – b | ≥ | c |, | b – c | ≥ | a |, |c-a|≥|b|, then prove that one of a ,b, c is the sum of the other two.

Answer 1

Explanation:

Using | a – b | ≥ | c |, we obtain |a – b|² ≥ |c|², which is equivalent to (a – b – c)(a – b + c) ≥ 0. Similarly, (b – c – a) (b – c + a) ≥ 0 and (c – a – b) (c – a + b) ≥ 0 .Multiplying these inequalities, We get 

- (a + b – c)² (b + c – a)² (c +a-b)² ≥ 0.

This forces that LHS is equal to zero. Hence it follows that either a + b = c, or b + c = a, or c + a = b