Norm inequality $\| x - y \| \cdot \| z \| \leq \| x - z \| \cdot \| y \| + \| z - y \| \cdot \| x \|$ 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..
 A: This is a condensed version of @Zvi's answer. The proof below would not have been possible without that answer. 
They key observation is the following identity:
$$
\left\| \frac{\|y\|}{\|x\|} x - \frac{\|x\|}{\|y\|} y \right \| = \|x-y\|,
$$
which can be proven by squaring.
Using this indentity, we get 
$$\begin{split}
\|x-y \|\cdot\|z\|& =\left\| \frac{\|y\|}{\|x\|} x - \frac{\|x\|}{\|y\|} y \right \|\|z\|
\\
&= \left\| \frac{\|y\|\|z\|}{\|x\|} x - \frac{\|x\|\|y\|}{\|z\|}z  + \frac{\|x\|\|y\|}{\|z\|}z  -\frac{\|x\|\|z\|}{\|y\|} y \right \| \\
&\le \left\| \frac{\|y\|\|z\|}{\|x\|} x - \frac{\|x\|\|y\|}{\|z\|}z \right \| + \left\| \frac{\|x\|\|y\|}{\|z\|}z  -\frac{\|x\|\|z\|}{\|y\|} y \right \| \\
&= \|y\|\left\| \frac{\|z\|}{\|x\|} x - \frac{\|x\|}{\|z\|}z \right \| + \|x\| \left\| \frac{\|y\|}{\|z\|}z  -\frac{\|z\|}{\|y\|} y \right \|\\
&= \|y\|\cdot \|x-z\| + \|x\| \cdot \|y-z\|.
\end{split}$$
A: If $x$, $y$, or $z$ is the zero vector, then the inequality is trivial.  If the dimension $n=\dim V$ is $1$, the problem is also trivial.  From now on, assume that $n\geq 2$ and $x,y,z\neq 0$, so the angles between these vectors are well-defined.
Let $a=\Vert x\Vert$, $b=\Vert y\Vert$, and $c=\Vert z\Vert$.  Suppose that $\alpha=\angle(y,z)$, $\beta=\angle(z,x)$, and $\gamma=\angle(x,y)$.  Clearly, we have $$\alpha,\beta,\gamma\in[0,\pi]\ \wedge\ \alpha+\beta+\gamma\leq 2\pi.$$
We also have triangle inequalities among the angles:
$$\beta+\gamma\geq \alpha,\ \gamma+\alpha\geq \beta,\ \wedge\ \alpha+\beta\geq \gamma.$$  In particular, we have 
$$\gamma\leq \min\{\alpha+\beta,2\pi-\alpha-\beta\}.$$
That is,
$$\frac{\gamma}{2} \leq \min\left\{\frac{\alpha+\beta}{2},\pi-\frac{\alpha+\beta}{2}\right\}.$$
Because $0\leq \alpha,\beta,\gamma\leq 2\pi$, we get
$$0\leq \sin\frac{\gamma}{2}\leq \sin\frac{\alpha+\beta}{2}.\tag{1}$$
Now,
$$\Vert x-y \Vert\cdot\Vert z \Vert=c\sqrt{(a-b)^2+4ab\sin^2\frac{\gamma}{2}}=\sqrt{(ca-bc)^2+4(ca)(bc)\sin^2\frac{\gamma}{2}}.$$
By (1), we get
$$\Vert x-y \Vert\cdot\Vert z \Vert\leq \sqrt{(ca-bc)^2+4(ca)(bc)\sin^2\frac{\alpha+\beta}{2}}.$$
We also have
$$\Vert y-z \Vert\cdot\Vert x \Vert=a\sqrt{(b-c)^2+4bc\sin^2\frac{\alpha}{2}}=\sqrt{(ab-ca)^2+4(ab)(ca)\sin^2\frac{\alpha}{2}}$$
and
$$\Vert z-x\Vert\cdot\Vert y\Vert = b\sqrt{(c-a)^2+4ca\sin^2\frac{\beta}{2}}=\sqrt{(bc-ab)^2+4(bc)(ab)\sin^2\frac{\beta}{2}}.$$
So, the required inequality is immediate, if we can show that
\begin{align}\sqrt{(ca-bc)^2+4(ca)(bc)\sin^2\frac{\alpha+\beta}{2}}&\leq \sqrt{(ab-ca)^2+4(ab)(ca)\sin^2\frac{\alpha}{2}}\\&\hphantom{12345}+\sqrt{(bc-ab)^2+4(bc)(ab)\sin^2\frac{\beta}{2}}.\end{align}
For convenience, let $p=bc$, $q=ca$, and $r=ab$.  We are to show that
\begin{align}\sqrt{(p-q)^2+4pq\sin^2\frac{\alpha+\beta}{2}}&\leq \sqrt{(q-r)^2+4qr\sin^2\frac{\alpha}{2}}+\sqrt{(r-p)^2+4rp\sin^2\frac{\beta}{2}}.\tag{2}\end{align}
Let $u,v,w\in\mathbb{R}^2$ be vectors with $\Vert u \Vert=p$, $\Vert v\Vert =q$, and $\Vert w\Vert =r$ such that $\angle(v,w)=\alpha$ and $\angle(w,u)=\beta$, so that $\angle(v,u)=\alpha+\beta$ if $\alpha+\beta\leq \pi$, or $\angle(v,u)=2\pi-\alpha-\beta$ if $\alpha+\beta\geq \pi$.  Then, we have
$$\Vert u-v \Vert \leq \Vert w-v\Vert + \Vert u-w\Vert\tag{3}$$
by the triangle inequality.  The inequality (3) is precisely (2), and the claim is proved.
A: 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):

A: Here is a rewritten form of daw's answer. Note that for any two scalars $p$ and $q$ and any two vectors $u$ and $v$ whose norms are equal, we have
$$
\|pu-qv\|_2=\|qu-pv\|_2\tag{$\ast$}
$$
because there exists a linear isometry $Q$ such that $Qu=v$ and $Qv=u$ (see also A symmetry identity involving norms).
Now, let $x=au,\ y=bv$ and $z=cw$ for some scalars $a,b,c$ and some unit vectors $u,v,w$. Then
\begin{aligned}
\|x-y\|\|z\|
&=\|\color{red}{a}cu-\color{red}{b}cv\|\\
&=\|\color{red}{b}cu-\color{red}{a}cv\|\ \text{ by $(\ast)$}\\
&\le\|b\color{red}{c}u-\color{red}{a}bw\| + \|a\color{red}{c}v-a\color{red}{b}w\|\\
&=\|\color{red}{a}bu-b\color{red}{c}w\| + \|a\color{red}{b}v-a\color{red}{c}w\|\ \text{ by $(\ast)$}\\
&=\|x-z\|\|y\|+\|y-z\|\|x\|.
\end{aligned}
