How to prove this inequality in Euclidean space? 
Prove that
  $$\begin{align*}&|a+b||a+c|+|a+b||b+c|+|a+c||b+c|\\
\leq &(|a|+|b|+|c|) \cdot |a+b+c|+|a||b|+|a||c|+|b||c|\end{align*}$$
  in Euclidean space $\mathbb{R}^n$.

I have been thinking about this inequality for 2 weeks. This exercise was in my exam in functional analysis. I think, we have to use fact, that
$|x+y|^2=\langle x,x\rangle+2\langle x,y\rangle +\langle y,y\rangle$ and $|x+y|\leq |x|+|y|$ and symmetries properties, but I can not find a good proof.
If $a,b,c\in \mathbb{R}$ then it is easy to prove this inequality. We have to prove following inequalities
$$|a+b||a+c|\leq |a||a+b+c|+|b||c|$$
$$|a+b||b+c|\leq |b||a+b+c|+|a||c|$$
$$|a+c||b+c|\leq |c||a+b+c|+|a||b|$$
There is symetry therofore we can prove only one equation. It is easy to show that
$$|a+b||a+c|=|a(a+b+c)+bc|\leq |a||a+b+c|+|b||c|$$
We take the sum of 3 inequalities above and the proof is ended.
I was trying to prove inequality analogues in the space $\mathbb{R^n}$, but without a success. If I take a square of one of the inequalities, I can not get simplifier inequality.
P.S. Please, correct my grammar mistakes
 A: One cheapish trick is to use the quaternions. 
For $a,b,c\in \mathbb{R}^n$, there exists a three dimensional subspace containing $a,b,c$. Since the inequality you wrote is obviously invariant under global isometries of $\mathbb{R}^n$, we can without loss of generality assume that $a\neq 0$ is real, and $a,b,c \in \mathbb{R}^4$ which we identify with the quaternions $\mathbb{H}$. 
The advantage to working in the quaternions is that it is an algebra, and has the property that 
$$ |pq| = |p||q| $$
Therefore we get
$$ |a+b||a+c| = |a^2 + ac + ba + bc| = |a(a+b+c) + bc| $$
here we see that it is important we choose $a$ to be real, so that $ba = ab$. 
Similarly
$$ |a+b||b+c| = |ab + ac + b^2 + bc| = |ba + b^2 + bc + ac| = |b(a+b+c) +ac | $$
and
$$ |a+c||b+c| = |ab + ac + cb + c^2| = |ab + ca + cb + c^2| = |c(a+b+c) + ab| $$
and you can apply directly your argument for the case $a,b,c$ are in $\mathbb{R}$ and argue that the desired inequality holds. Note that this also gives you when the inequality is in fact an equality: whenever $bc$ and $a(a+b+c)$ are positively collinear, $ac$ and $b(a+b+c)$ are positively collinear, and $ab$ and $c(a+b+c)$ are positively collinear as quaternions. The trivial cases are when $a,b,c$ are all collinear and all have the same sign, and when $a+b+c = 0$. 
