# 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

-
Hi, I see that you are new to math.stackexchange! Since this is your first post, I would like to say that in general when you post a question you should just throw it in and ask people to prove/solve/do it. You should include at least some of your thoughts on how to attack the problem, or if you have no ideas say what you are having difficulty with. –  fpqc Dec 19 '11 at 12:10
I think Benjamin left out a little word. You should not just throw it in, but instead should include your own thoughts and the source of the problem. –  mixedmath Dec 19 '11 at 12:12
Here's some hints as to how I would try to prove this. You will probably use the triangle inequality $|x+y| \leq |x| + |y|$ a lot. And since you see $|a+b+c|$ on the right side and not on the left, I would try things like $|a+b| =|a+b+c-c| \leq |a+b+c|+|c|$. Just play around with it a bit. –  Jeff Dec 20 '11 at 14:28
@Jeff I think that approach is unlikely to get the inequality. Note that the inequality is tight when we plug in $a = b = c = 1$, so it means that every intermediate bound should be tight as well. But $|a+b| \leqslant |a+b+c|+|c|$ is too wasteful (LHS=2, RHS=4). –  Srivatsan Dec 21 '11 at 9:14

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 $a+b+c = 0$.

-