If we assume Cauchy-swarz $|\vec{u}||\vec{v}|\ge |\vec{u}\cdot\vec{v}|$, is the following proof for triangle inequality correct?
$$|\vec{u}+\vec{v}|^2 = |\vec{u}|^2+|\vec{v}|^2+2(\vec{u}\cdot\vec{v})$$ $$(|\vec{u}|+|\vec{v}|)^2=|\vec{u}|^2+|\vec{v}|^2+2|\vec{u}||\vec{v}|$$ since $|\vec{u}||\vec{v}|\ge |\vec{u}\cdot\vec{v}|\Rightarrow |\vec{u}||\vec{v}|\ge (\vec{u}\cdot\vec{v})$ $$|\vec{u}|^2+|\vec{v}|^2+2(\vec{u}\cdot\vec{v})\le |\vec{u}|^2+|\vec{v}|^2+2|\vec{u}||\vec{v}|$$ $$\Leftrightarrow |\vec{u}+\vec{v}|^2\le (|\vec{u}|+|\vec{v}|)^2$$ $$\Leftrightarrow |\vec{u}+\vec{v}|\le (|\vec{u}|+|\vec{v}|)$$