Answer
See the explanation below.
Work Step by Step
Let $u=(1,0,2), \quad v=(1,2,0)$, $\langle u, v \rangle=u\cdot v$
Then, we have
$\langle u, v \rangle =u_1v_1+u_2v_2+u_3v_3=1+0+0=1$, $\| u \| =\sqrt{\langle u, u\rangle}=\sqrt{u_1^2+u_2^2+u_3^2}=\sqrt{1+0+4}=\sqrt{5}$, $\| v \| =\sqrt{\langle v, v\rangle}=\sqrt{v_1^2+v_2^2+v_3^2}=\sqrt{1+4+0}=\sqrt{5}$, $\| u+v \| =\| (2,2,2) \| =\sqrt{4+4+4}=\sqrt{12}$
Now, we get
(a) Cauchy-Schwarz Inequality:
$|\langle u, v \rangle|=1\leq\| u \|\| v \|=\sqrt 5\sqrt 5 =5$
(b) The triangle inequality:
$\| u+v \| =\sqrt{12}=3.46\leq\| u \|+\| v \|=\sqrt {5}+\sqrt {5}=4.47$.
