Let $f\in\mathbb{R}^n$ and $g\in\mathbb{R}^n$ be two orthogonal unit vectors such that $\sum_{i}{f_i}=\sum_{i}{g_i}=0$.

Question. Can we prove this? $$\frac{\sum_{\{i,j\}}\min((f_i-f_j)^2,(g_i-g_j)^2)}{\sum_{\{i,j\}}\max((f_i-f_j)^2,(g_i-g_j)^2)} \ge \frac{1}{2n-1}$$

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    $\begingroup$ Just a thought. Unit vectors summing to 0 satisfy $\sum_{i,j} (f_i - f_j)^2 = 2n$, so your claim is equivalent to showing $\sum_{i,j} \max( (f_i - f_j)^2, (g_i - g_j)^2) \leq 2(2n-1)$. Do you have any examples with this bound being tight? $\endgroup$
    – Pat Devlin
    Jan 21, 2017 at 17:15
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    $\begingroup$ @PatDevlin: In addition to the Amdeberhan example, there are many other tight examples. $\endgroup$
    – j.s.
    Jan 22, 2017 at 17:23
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    $\begingroup$ An equivalent statement is $\sum_{i,j}|(f_i-f_j)(g_i-g_j)|\le 2n-2$, in the same assumptions on $f$, $g$. Note that Cauchy-Schwarz almost does it: $\sum_{i,j}|(f_i-f_j)(g_i-g_j)|\le 2n$ $\endgroup$ Feb 2, 2017 at 17:08
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    $\begingroup$ 1) Use $2\max(a,b)=a+b+|a-b|$ and $2\min(a,b)=a+b-|a-b|$ in any $\max$ and $\min$ ; 2) Replace $(f,g)$ with $({f+g \over {\sqrt 2}},{f-g\over {\sqrt 2}})$: this is a bijection of the minimization domain into itself. $\endgroup$ Feb 3, 2017 at 9:21
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    $\begingroup$ About $(f,g)\mapsto(f',g'):=({f+g\over \sqrt{2}}, {f-g\over \sqrt{2}})$: one just notes it is an involution of the set described in the first line of the OP. $\endgroup$ Feb 3, 2017 at 9:55

1 Answer 1


The example below shows that the bounds are tight.

Take the vectors $v=[1,2,\dots,(n-2),-\binom{n-1}2,0]$ and $w=[-1,-1,\dots,-1,(n-1)]$, then normalize them, with $\Vert v\Vert^2=\frac{3n-5}2\binom{n}3$ and $\Vert w\Vert^2=n(n-1)$, to get $$f=\frac{v}{\Vert v\Vert}\qquad \text{and} \qquad g=\frac{w}{\Vert w\Vert}.$$ Then, $\sum_{i,j}\max_{i,j}\{(f_i-f_j)^2,(g_i-g_j)^2\}$ equals \begin{align} 2\sum_{1\leq i<j\leq n-2}\frac{(i-j)^2}{\Vert v\Vert^2}+ 2\sum_{i=1}^{n-2}\frac{(i+\binom{n-1}2)^2}{\Vert v\Vert^2}+2\sum_{i=1}^{n-1}\frac{n^2}{\Vert w\Vert^2}=2(2n-1). \end{align}


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