For inner product spaces, do we have $||\vec{u}-\vec{v}|| \leq ||\vec{u}||+||\vec{v}||$?

Question

Let $V$ be an inner product space. Then for all $\vec{u},\vec{v} \in V$ we have $$||\vec{u}-\vec{v}|| \leq ||\vec{u}||+||\vec{v}||.$$

I know that the converse to the equation is true such that $||\vec{u}+\vec{v}||$ is less than or equal to $||\vec{u}||+||\vec{v}||$. Is this also true?

Answer 1

Take the triangle inequality $\|x + y\| \le \|x\| + \|y\|$. What happens if you do $y\leftarrow -y$?

Answer 2

It's true not only in inner-product spaces but also in normed vector spaces in general (in fact, it is sufficient that $\|\cdot\|$ is merely a seminorm on $V$). As mentioned by @ncmathsadist, use the triangle inequality and absolute homogeneity of the norm: $\|\lambda\mathbf v\|=|\lambda|\|\mathbf v\|$ for any $\mathbf v\in V$ and any $\lambda$ in the underlying field (in particular, $\lambda=-1$, if we're talking about real or complex vector spaces).