# Triangle Inequality and Its Proof

Triangle Inequality states that for any real numbers a and b,

|a| + |b| ≥ |a + b|.

In proving this theorem, we use the definition of absolute value of a number. The absolute value of a number is x if x > 0 (*), –x  if x < 0 (**) and 0 if x = 0 (**). For instance, the absolute value of 5 is 5 since 5 > 0. On the other hand, the absolute value of -3 is -(-3) = 3. Clearly, the absolute value of 0 is 0. The geometric representation of triangle inequality is shown above (see Demystifying Triangle Inequality). It states that a triangle can only be formed if the combined length of any two sides is greater than that of the third sides. Using the definition of absolute value, we now prove the theorem. Theorem For any real numbers a and b, |a| + |b| ≥ |a + b|.

Proof

Case 1: a and b are positive If a and b are both positive, then a + b and |a + b| are also positive. From *,

|a| + |b| = a + b = |a + b|.

Case 2 : a and b are both negative If a and b are both negative, from **, |a| + |b| = (-a) + (-b) = -(a + b) But from **, |x| = –x if x <0, so -(a  + b) = |a + b| which means that

|a| + |b| = |a + b|.

Case 3: a is positive and b is negative |a| + |b| = a + (-b) and |a + b| is a + b itself (if positive) or –(a + b) if negative. But, a > –a and – b  > b (why?) so,

a + (-b) > a  + b  and (-a) + (-b) > -(a + b).

In either case, we have |a| + |b| > |a + b|. Therefore, combining the three cases, we have

|a| + |b| ≥ |a + b|.