Question

Prove that |a+b|≤|a|+|b| where a,b are real numbers

Answer 1

We can start by considering two cases:

Case 1: a+b ≥ 0

In this case, we have

|a+b| = a+b

|a|+|b| = a+b

So it is clear that |a+b| ≤ |a|+|b|.

Case 2: a+b < 0

In this case, we have

|a+b| = -(a+b)

|a|+|b| = a+(-b), since |b| = -b in this case

Adding the two inequalities gives

|a+b| + (|a|+|b|) ≤ a+b + (a-b) = 2a ≤ 2|a|

Therefore, we have

|a+b| ≤ 2|a| - |b|

It remains to show that 2|a| - |b| is nonnegative. Since |a| is nonnegative, we have

2|a| - |b| ≥ 2|a| - |a| = |a| ≥ 0

Thus, we have shown that |a+b| ≤ |a|+|b| in both cases, and the result follows.