Answer
(a) See the attached figure for a geometric interpretation.
(b) $ |a + b|\leq |a|+ |b|$
Work Step by Step
(a) As we are given: $ |a + b|\leq |a|+ |b|$
Therefore, the given inequality tells that the length of one side of a triangle is less than the sum of the lengths of the other two sides.
See the attached figure for a geometric interpretation.
(b) By contradiction let us assume that $ |a + b| \gt |a|+ |b|$
$ |a + b| ^2\gt( |a|+ |b|)^2$
$ (a+b) \cdot (a+b)= a .a+a.b+b.a+b.b=|a|^2+2(a \cdot b)+|b|^2$
$a \cdot b \gt |a||b|$
If $a \cdot b \gt 0$ then $|a \cdot b| a \cdot b$, so $ |a \cdot b|\gt |a| |b|$
This contradicts the Cauchy-Schwarz Inequality.
Therefore, the contradiction has been proven wrong.
Hence, $ |a + b|\leq |a|+ |b|$
