So, let us assume that |
|
0
|
|
a
b
≠
≠
r
r Then, we have
|
|
|
||
a b|
a b
⋅
rr
r
r
 = | cos |
1
θ ≤
Therefore
|
a b|
⋅
rr
≤ |
||
|
a
rb
r
Example 20 For any two vectors 
aand
rb
r
, we always
have |
|
|
|
|
|
a
b
a
b
+
≤
+
r
r
r
r
(triangle inequality) Solution The inequality holds trivially in case either
0 or
0
a
b
=
r=
r
r
r
 (How )