Prove that triangle inequality holds for some function

I have the following problem (considering the vector space structure of $\mathbb{R}^n$):

Let $d: \mathbb{R}^n \times \mathbb{R}^n \to [0,\infty)$ be defined by: $$d(x,y)=\begin{cases}\| x-y \|, \mbox{if x and y are linearly dependent}\\ \|x\|+\|y\|, \mbox{otherwise} \end{cases}$$

Prove that $(\mathbb{R}^n,d)$ is a metric space.

I had no difficulties prooving that $d$ satisfies the propoerties of a metric function but the triangle inequality one.

For proving that $\forall x,y,z \in \mathbb{R}^n$ the triangle inequality holds, I divided by cases:

1. $x,y$ are linearly independent but $x,z$ not;
2. $x,y$ L.D. and $x,z$ L.D.;
3. $x,y$ L.I but $x,z$ L.D.; and
4. $x,y,z$ L.I.

I could prove all except 3. I have

$d(x,y)=\|x\|+\|y\|$, and $d(x,z)+d(z,y)=\|x-z\|+\|z\|+\|y\|$

I tried to use that $\vert \|a\|-\|b\| \vert \leq \|a+b\|$, but I couldn't...

Could you help me? Thanks in advance.

OBS.: I can't find a basic proof here to: prooving that the following are metrics (provided that $d$ is) $d_1(x,y)=\min\{1,d(x,y)\}; d_2(x,y)=\frac{d(x,y)}{1+d(x,y)}$; (I got stuck prooving triangle inequality too)

• What does "L.D." mean in this context? – kimchi lover Aug 27 '18 at 21:32
• @kimchilover I'd guess at linearly dependent – Tom Collinge Aug 27 '18 at 22:02
• Linearly dependent! I'm sorry.. – Robson Aug 27 '18 at 22:46

For (3), if $x, z$ are linearly dependent then for some $\alpha$, $z = \alpha x$
And if $x, y$ are linearly independent then so are $z, y$
So, $d(x, z) + d(z, y) = ||x - z|| + ||z|| + ||y||$
$= ||(1 - \alpha) x|| + ||z|| + ||y|| = |1-\alpha|.||x|| + ||\alpha x|| + ||y||$
$= (|1 -\alpha| + |\alpha|) ||x|| + ||y||$
Which for any value of $\alpha$ is $\ge ||x|| + ||y|| = d(x, y)$