Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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. – user38268 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 $ba = ab$.


$$ |a+b||b+c| = |ab + ac + b^2 + bc| = |ba + b^2 + bc + ac| = |b(a+b+c) +ac | $$


$$ |a+c||b+c| = |ab + ac + cb + c^2| = |ab + ca + cb + c^2| = |c(a+b+c) + ab| $$

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 $a+b+c = 0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.