prove that a + b ≤ |a| + |b|?

|a| ≥ O and |b| ≥ 0

Case I: a ≥ 0  and  b ≥ 0
When a ≥ 0:  |a| = a …… [1]
When b ≥ 0:  |a| = b …… [2]
[1] + [2]:  |a| + |b| = a + b

Case II: a ≥ 0  and  b < 0
When a ≥ 0:  |a| = a …… [3]
When b < 0:  |b| > b …… [4]
[3] + [4]:  |a| + |b| > a + b

Case III:
a < 0  and  b ≥ 0
When a < 0:  |a| > a …… [5]
When b ≥ 0:  |a| = b …… [6]
[5] + [6]:  |a| + |b| > a + b

Case IV:
a < 0  and  b < 0
When a < 0:  |a| > a …… [7]
When b < 0:  |a| > b …… [8]
[7] + [8]:  |a| + |b| > a + b

Combine the above cases.  Conclusion: |a| + |b| ≥ a + b

You can easily prove these two statements, right?
a ≤ |a|
b ≤ |b|

For example:
-3 < |-3|
0 = |0|
5 = |5|

So from there it's a simple step to get to:
a + b ≤ |a| + |b|

You have done the resultant  R of two vectors
R^2= a2 +b^2 +2ab cos theta=||a||^2+||b||^2+ 2 ||a|| ||b|| cos theta
a+b = |a| +|b| only when cos theta=0 or theta=pi/2.. Otherwise a+b > |a|+|b|

Absolute values are never negative. Thus the sum of |a| and |b| is > than a + b if one or both are negative.  Equality holds if both are non negative.