Answer
See explanation.
Work Step by Step
Let $\vec{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle$ , $\vec{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle$ be vectors of $\mathcal{R}^{3}$
\[
\begin{array}{rl}
\vec{v}+\vec{u} & =\left\langle v_{1}+u_{1}, v_{2}+u_{2}, v_{3}+u_{3}\right\rangle \\
\Rightarrow\|\vec{v}+\vec{u}\| & =\sqrt{\left(v_{1}+u_{1}\right)^{2}+\left(v_{2}+u_{2}\right)^{2}+\left(v_{3}+u_{3}\right)^{2}} \\
\Rightarrow\|\vec{v}+\vec{u}\|^{2} & =\left(u_{1}^{2}+u_{2}^{2}+v_{3}^{2}\right)+\left(v_{1}^{2}+v_{2}^{2}+v_{3}^{2}\right)+2\left(u_{1} v_{1}+u_{2} v_{2}+u_{3} v_{3}\right) \\
& =\|\vec{u}\|^{2}+\|\vec{v}\|^{2}+2 \vec{u} \cdot \vec{v} \\
& =[\|\vec{u}\|+\|\vec{v}\|]^{2}+2(\|\vec{u}\|\|\vec{v}\| \cos (\theta)-\|\vec{u}\|\|\vec{v}\|) \\
& =[\|\vec{u}\|+\|\vec{v}\|]^{2}-2(1-\cos (\theta))\|\vec{u}\|\|\vec{v}\| \\
\operatorname{since}|\cos (\theta)| \leq 1 & \operatorname{and}\|\vec{u}\|,\|\vec{v}\| \geq \\
& \|\vec{u}+\vec{v}\|^{2} \leq\left(\|\vec{u}\|^{2}+\|\vec{v}\|^{2}\right) \\
& =[|\vec{u}+\vec{v}||\leq||\vec{u}||+||\vec{v}| |
\end{array}
\]
For example, if $\vec{u}=\vec{v}$
\[
\|\vec{u}+\vec{v}\|=\|2 \vec{u}\|=2\|\vec{u}\|=\|\vec{u}\|+\|\vec{u}\|=\|\vec{u}\|+\|\vec{v}\|
\]
For $\vec{v}=\langle 0,1\rangle$ and $\vec{u}=\langle 1,0\rangle$ where $\|\vec{v}\|=\|\vec{u}\|=1$
\[
\|\vec{v}+\vec{u}\|=\|\langle 1,1\rangle\|=\sqrt{1+1}=\sqrt{2}<2=1+1=\|\vec{v}\|+\|\vec{u}\|
\]
