Explanation for the proof of Ptolemy-inequality

Question

In the book of linear algebra by Werner Greub, at page $190$, it is given the proof of the Ptolemy-inequality,

$$|x-y| |z| \leq |y-z| |x| + |z-x||y|,$$ where $x,y,z$ are in a real inner product space, and |.| is the Euclidian norm.

So how can we prove that if at least one of the vectors is zero, then the inequality holds ?

In case you want to see the complete proof,

Edit:

Right now, I only need to prove $|x||z| = |xz|$.

Answer 1

I mainly asked this question because I thought the proof was incomplete, but thanks to @ama's comment, I realised what I have read wrong, so for the rest of the problem, i.e at least one of the vectors is zero case,

Since $|- a| = |a|$ by definition, when we assume one of them is zero, the inequality leaves it place to equality, so the statement holds.

Answer 2

If the variables are scalar (not vectors), then: $$|a+b|\le |a|+|b|.$$ Thus: $$|x-y|=|x-z+z-y|\le |x-z|+|z-y|,$$ $$|x-y||z|=|xz-yz|=|xy-zx+zy-xy|\le |xy-zx|+|zy-xy|=|y-z||x|+|z-x||y|.$$