Answer
False.
Work Step by Step
Counter example:
a = <1,-1,1>
b = <1,1,-1>
|a| = $\sqrt{1^2+1^2+1^2}$ = $\sqrt{3}$
|b| = $\sqrt{1^2+1^2+1^2}$ = $\sqrt{3}$
|a| + |b| = $\sqrt{3}$ + $\sqrt{3}$ = $\mathbf{2\sqrt{3}}$
a+b = <2,0,0>
|a+b| = $\sqrt{2^2+0+0}$ = $\sqrt{4}$ = $\mathbf{2}$
2$\sqrt 3$ $\ne$ 2
False.
