For any two vectors $\vec{a}$ and $\vec{b},$ we always have $|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|$ (triangle inequality).

(N/A) The inequality holds trivially in case either $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0}.$ So,let $|\vec{a}| \neq 0$ and $|\vec{b}| \neq 0.$ Then,
$|\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b})$
$= \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b}$
$= |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2$ (since the scalar product is commutative)
$\leq |\vec{a}|^2 + 2|\vec{a} \cdot \vec{b}| + |\vec{b}|^2$ (since $x \leq |x|$ for all $x \in \mathbb{R}$)
$\leq |\vec{a}|^2 + 2|\vec{a}||\vec{b}| + |\vec{b}|^2$ (by Cauchy-Schwarz inequality,$|\vec{a} \cdot \vec{b}| \leq |\vec{a}||\vec{b}|$)
$= (|\vec{a}| + |\vec{b}|)^2$
Taking the square root on both sides,we get:
$|\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}|$