Inequality related to $\sum_{i=1}^{m}\sum_{j=i+1}^{m} \left\Vert a_{i}+a_{j}\right\Vert$

Question

Given vectors $a_{i} \in \mathbb R^3$ ,where $i$ is a natural number satisfying $1\le i\le 3$, then the following relation does hold: $$\left\Vert a_1+a_2\right\Vert+\left\Vert a_1+a_3\right\Vert+\left\Vert a_2+a_3\right\Vert\le\left\Vert a_1\right\Vert+\left\Vert a\right\Vert+\left\Vert a_3\right\Vert+\left\Vert a_1+a_2+a_3\right\Vert$$

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I Tried to generalized the problem: Given vectors $a_{i} \in \mathbb R^n$ ,where $i$ is a natural number satisfying $1\le i\le m$,then the following relation does hold:

$$\sum_{i=1}^{m}\sum_{j=i+1}^{m} \left\Vert a_{i}+a_{j}\right\Vert \le\sum_{i=1}^{m} \left\Vert a_{i} \right\Vert+ \left\Vert \sum_{i=1}^{m}a_{i}\right\Vert$$

Is the generalization true?if yes then how to prove that?

Answer 1

No. Let $a_i=a\neq 0$ for all $i$. Then LHS is $m(m-1)\lVert a\rVert$ and RHS is $2m\lVert a\rVert$, so LHS is larger than RHS for sufficiently large $m$.

Answer 2

This is known as Hlawka's inequality. It can be written as $$\mathbb E\left\|\sum_{i=1}^3\epsilon_ix_i\right\|\ge\frac{1}{2}\sum_{i=1}^3\|x_i\|$$ where $x_i=\frac{a_i+a_{i+1}}{2}$ (with $a_4=a_1$) and the $\epsilon_i$s are i.i.d. random variables such that $\epsilon_i=\pm1$ with probability $\frac12$. According to Honda et al., Generalizations of the Hlawka's Inequality, Bull. Kyushu Inst. Tech., Pure Appl. Math., no. 45, 1998, pp.9-15, it has the generalization $$\mathbb E\left\|\sum_{i=1}^n\epsilon_ix_i\right\|\ge\frac{1}{2^{n-1}}\binom{n-1}{\lfloor\frac{n}{2}\rfloor}\sum_{i=1}^n\|x_i\|$$ (where $x_1,x_2,\ldots,x_n$ are any $n$ vectors in an inner product space) and this inequality is tight. When $n=4$, if I am not mistaken, by setting $x_i=\frac{a_i+a_{i+1}}{2}$ (with $a_{n+1}=a_1$) the inequality can be simplified to \begin{aligned} &\|a_1+a_2\|+\|a_2+a_3\|+\|a_3+a_4\|+\|a_4+a_1\|\\ \le\ &2\left(\|a_1+a_2+a_3+a_4\|+\|a_1-a_3\|+\|a_2-a_4\|\right). \end{aligned}