TITLE: Norm inequality $\| x - y \| \cdot \| z \| \leq \| x - z \| \cdot \| y \| + \| z - y \| \cdot \| x \|$
QUESTION [9 upvotes]: Let $x$, $y$, $z$ be $3$ vectors in Euclidean space $V$.
$\| x \|$ is norm of $x$ (length)
How do you prove:
$$\| x - y \| \cdot \|z\| \leq \| x - z\| \cdot \| y \| + \| z - y\| \cdot \| x \|?$$
I have tried Cauchy-Schwartz inequality, | < x,y > | <= ||x|| •||y||, where  is dot product between x and y, trying to move everything on one side and then proving that it is greater than zero. Also I used modules properties, |a + b| < = |a| + |b| and so on.
Also triangle inequality, ||x + y || <= ||x|| + ||y|| , trying to get the other side of inequality.
Considering that ||x|| is sqrt( < x,x > ) and < x , a+b > is < x ,a > + < x ,b > and that, norm is always greater than 0 and sqrt( a+b )<= sqrt(a) + sqrt(b) , if a and b are positive, I played with this properties without getting any result..
Also distance formula, d(a,b) <= d(a,c) + d(c,a), where d(a,b) = || a - b||
There is this formula: Ilz - xII^2 + Ilz - yII^2 - Ilx - yII^2 =2 Maybe it helps. I have tried to use it but without a result..

REPLY [3 votes]: You wait all day for a rewrite, and then three of them come along at once! :)
The inequality is clearly true if any of $x, y, z$ is zero, so we can suppose that none of them are.
The inequality then becomes:
$$
d(x, y) \leqslant d(x, z) + d(z, y)
$$
where:
$$
d(x, y) = \frac{\|x - y\|}{\|x\|\|y\|} \ (x \ne 0, \ y \ne 0), \text{ etc.}
$$
But we can write:

$$
d(x, y) = \|Tx - Ty\|,
$$

where $T$ is a kind of "inversion":
$$
T \colon V \setminus \{0\} \to V \setminus \{0\}, \ x \mapsto \frac{x}{\|x\|^2}.
$$
Proof. As daw says, it's just a matter of squaring:
$$
d(x, y)^2 = \frac{\left\langle x - y, x - y \right\rangle}{\|x\|^2\|y\|^2} =
\frac{1}{\|y\|^2} - \frac{2\left\langle x, y \right\rangle}{\|x\|^2\|y\|^2} + \frac{1}{\|x\|^2} = \|Ty\|^2 - 2\left\langle Tx, Ty \right\rangle + \|Tx\|^2,
$$
i.e. $d(x, y)^2 = \|Tx - Ty\|^2$. $\square$
And now we have:
$$d(x, y) = \|Tx - Ty\| \leqslant \|Tx - Tz\| + \|Tz - Ty\| = d(x, z) + d(z, y),
$$
as required. $\square$

From p.185 of Werner H. Greub, Linear Algebra (3rd ed. 1967) (see comments on daw's answer):