Prove that |a + b| ≤ |a| + |b|.
|a| = √a2
The Attempt at a Solution
Since |a| = √a2, then
|a + b| = √(a + b)2 = √(a2 + 2ab + b2) = √a2 + √b2 + √(2ab) = |a| + |b| + √(2ab).
And since the square root of a negative number is not defined, then 2ab must be ≥ 0.
This proves the theorem that |a + b| ≤ |a| + |b|.