How to prove this inequality in Euclidean space?

Solution 1:

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 $a+b+c = 0$.