How prove this $|a-b||c|\le|a-c||b|+|b-c||a|$ if $a,b,c$ are vectors? Let $a,b,c$ are vector numbers,show that
$$|a-b||c|\le|a-c||b|+|b-c||a|$$
my idea: $a_{i},b_{i},c_{i}\in \mathbb{R}$
let
$$a=(a_{1},a_{2},\cdots,a_{n})$$
$$b=(b_{1},b_{2},\cdots,b_{n})$$
$$c=(c_{1},c_{2},\cdots,c_{n})$$
$$\Longleftrightarrow \sqrt{\sum_{i=1}^{n}(a_{i}-b_{i})^2}\sqrt{\sum_{i=1}^{n}c^2_{i}}\le\sqrt{\sum_{i=1}^{n}(a_{i}-c_{i})^2}\sqrt{\sum_{i=1}^{n}b^2_{i}}+\sqrt{\sum_{i=1}^{n}(b_{i}-c_{i})^2}\sqrt{\sum_{i=1}^{n}a^2_{i}}$$
I think follow is use Cauchy-Schwarz or Minkowski inequality. But I can't, I feel this inequality is interesting.
 A: This can be viewed as an inequality about simplices in ${\mathbb R}^3$, as only the four points $a$, $b$, $c$, and $O$ are involved. It then says that the product of the length of any two opposite edges is at most equal to the sum of the two other such products. 
In order to prove this we may assume
$$O=(0,0,0),\quad a=(a,0,0), \quad b=(b_1,b_2,h),\quad c=(c_1,c_2, h)\ ,$$
i.e., that $b\vee c$ is parallel to the plane $z=0$. We then have to prove that
$$\eqalign{|Oa||b-c|&\leq |Ob||a-c|+|Oc||a-b|\cr &
=\sqrt{|Ob'|^2+h^2}\sqrt{|a-c'|^2+h^2}+\sqrt{|Oc'|^2+h^2}\sqrt{|a-b'|^2+h^2}\ ,
\cr}\tag{1}$$
where we have put $(b_1,b_2,0)=:b'$, $\>(c_1,c_2,0)=:c'$. Note that $|b-c|=|b'-c'|$; therefore the left hand side of $(1)$ is independent of $h$. It follows that it is enough to prove $(1)$ for $h=0$, in which case it can be written as
$$|Oa||b'-c'|\leq |Ob'||a-c'|+|Oc'||a-b'|\ .$$
Now this is Euler's generalization of Ptolemy's equality about cyclic quadrilaterals to an inequality about arbitrary convex quadrilaterals, see here.
