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Show that for all real numbers $x_1, x_2, \ldots, x_n$, $$ \sum_{i=1}^n \sum_{j=1}^n\left|x_i+x_j\right| \geq n \sum_{i=1}^n\left|x_i\right| $$

If we fix $x_2,.....x_n$, and vary $x_1$, then we can define $f(x_1)= |x_1|+c$ where $c=|x_2|+.....+|x_n|$, then we see the function is convex. Thus it achives it's maximum value at one of the endpoints on the interval, but in this case there's no endpoints, thus $x_1$ can then get arbitrarily big, then how to proceed?

math_learner
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  • start with small $n$'s. $n=1$ and $2$ look OK. Can you do $n=3$? if so do $n=4$. Draw some pictures. Do some tests for larger $n$. Is the constant on RHS the best possible ( here we are comparing some norms on a f. d. vector space, even for a fixed $n$ it may not be easy). Good luck! – orangeskid Jul 20 '24 at 19:49

1 Answers1

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Let $a_1,\cdots,a_r\geqslant0$ and $b_1,\cdots,b_s\geqslant0$ such that $r+s=n$ and $\{x_1,\cdots,x_n\}=\{a_1,\cdots,a_r\}\cup\{-b_1,\cdots,-b_s\}$.

Let $A=a_1+\cdots+a_r$ and $B=b_1+\cdots+b_s$.

We have to prove that :

$$\sum_{1\leqslant i,j\leqslant r}(a_i+a_j)+\sum_{1\leqslant i,j\leqslant s}(b_i+b_j)+2\sum_{i=1}^r\sum_{j=1}^s\vert a_i-b_j\vert\geqslant n(A+B)$$

ie :

$$2\sum_{i=1}^r\sum_{j=1}^s\vert a_i-b_j\vert\geqslant(s-r)(A-B)\tag{$\star$}$$

Four cases arise, depending on the signs of the differences $s-r$ and $A-B$. If those signs are opposite, then $(\star)$ is obvious. Let us assume that $s\geqslant r$ and $A\geqslant B$. We can see that :

$$2\sum_{i=1}^r\sum_{j=1}^s\vert a_i-b_j\vert\geqslant2\sum_{i=1}^r\sum_{j=1}^s(a_i-b_j)=2(sA-rB)$$ and this last quantity is greater than $(s-r)(A-B)$ because : $$\begin{eqnarray}2(sA-rB)-(s-r)(A-B)&=&(s+r)A+(s-3r)B\\ &\geqslant&(s+r)B+(s-2r)B\\ &=&2(s-r)B\\ &\geqslant&0\end{eqnarray}$$

The last case ($s\leqslant r$ and $A\leqslant B$) is similar.

Adren
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