Answer
See the explanation below.
Work Step by Step
$u=(0,3,\frac{1}{3}), \quad v=(\frac{4}{3}, 1,-3)$, $\langle u, v \rangle =2u_1v_1+ u_2v_2+2u_3v_3 $.
$$ \langle u, v\rangle =2u_1v_1+ u_2v_2+2u_3v_3=0+3-2=1,$$
$$\| u \|=\sqrt{\langle u, u\rangle} =\sqrt{2u^2_1+ u^2_2+2u^2_3}=\sqrt{0+9+\frac{2}{9}}=\sqrt{\frac{83}{9}}$$
$$\|v \|=\sqrt{\langle v, v\rangle} =\sqrt{2v^2_1+ v^2_2+2v^2_3}=\sqrt{\frac{32}{9}+1+18}=\sqrt{\frac{203}{9}}$$
Now, we have
The Cauchy-Schwarz Inequality: $|\langle u, v \rangle|=1\leq\| u \|\| v \|=\sqrt{\frac{83}{9}}\sqrt{\frac{203}{9}}=14.42$
The triangle inequality: $\| u+v \| =\| (\frac{4}{3},4,-\frac{8}{3}) \| =\sqrt{\frac{32}{9}+16+\frac{128 }{9}}=\sqrt {\frac{304}{9}}=5.81\\ \leq\| u \|+\| v \|=\sqrt{\frac{83}{9}}+\sqrt{\frac{203}{9}}=7.78.$
