Problem 4165 from Crux Mathematicorum

Problem

Problem 4165 from Crux Mathematicorum

Solution

We prove the stronger result that for any complex numbers $x_1,$ $x_2,$ $x_3$ and $x_4,$ we have

(1)

$\displaystyle \left|\sum_{i=1}^4x_i\right|+2\sum_{i=1}^4|x_i|\ge \sum_{1\le i\lt j\le 4}|x_i+x_j|.$

The proposed inequality then follows from (1) by the AM-GM inequality. To prove (1), we will make use of Hlawka's inequality which states that

(2)

$|a+b+c|+|a|+|b|+|c|\ge |a+b|+|b+c|+|c+a|$

for all complex numbers $a,b,c.$ Setting $a=x_1,\,$ $b=x_2,\,$ and $c=x_3+x_4,\,$ we have from (2)

(3)

$\displaystyle\begin{align}&|x_1+x_2+x_3+x_4|+|x_1|+|x_2|+|x_3+x_4|\\ &\qquad\qquad\ge |x_1+x_2|+|x_2+x_3+x_4|+|x_1+x_3+x_4|.\end{align}$

Applying (2) again, we obtain

(4)

$\displaystyle\begin{align}&|x_2+x_3+x_4|\\ &\qquad\qquad\ge |x_2+x_3|+|x_3+x_4|+|x_4+x_2|-|x_2|-|x_3|-|x_4| \end{align}$

and

(5)

$\displaystyle\begin{align}&|x_1+x_3+x_4|\\ &\qquad\qquad\ge |x_1+x_3|+|x_3+x_4|+|x_4+x_1|-|x_1|-|x_3|-|x_4|. \end{align}$

Adding (4) and (5) and denoting the right side of (3) by $R\,$ gives

(6)

$\displaystyle\begin{align}R &\ge |x_3+x_4|-|x_1|-|x_2|-2|x_3|-2|x_4|+\sum_{1\le i\lt j\le 4}|x_i+x_j|. \end{align}$

From (3) and (6), we deduce that

$\displaystyle\begin{align}&|x_1+x_2+x_3+x_4|+|x_1|+|x_2|+|x_3+x_4|\\ &\qquad\qquad\ge |x_3+x_4|-|x_1|-|x_2|-2|x_3|-2|x_4|+\sum_{1\le i\lt j\le 4}|x_i+x_j| \end{align}$

from which (2) follows immediately.

Generalization

In a normed space, Hlawka-like inequality

$||x+y+z||+||x||+||y||+||z||\ge ||x+y||+||y+z||+||z+x||$

implies a more general one

$\displaystyle (n-2)\sum_{k=1}^n||x_k||+\left|\left|\sum_{k=1}^nx_k\right|\right|\ge\frac{n(n-1)}{2}\left(\prod_{1\le i\lt j\le n}||x_i+x_j||\right)^{\frac{2}{n(n-1)}}.$

Acknowledgment

This is problem 4165 from the Canadian Crux Mathematicorum (Vol. 43(7), September 2017); the problem was posed by Dan Sitaru, the solution is by Michel Bataille, the generalization is by Leo Giugiuc. Leo refers to the generalization as the Adamovic-Savic inequality.

 

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