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Say that a vector $x=(x_1,x_2, \ldots ,x_n)\in {\mathbb R}^n$ is nondecreasing if $x_1 \leq x_2\leq \ldots \leq x_n$. Can anyone show or find a counterexample to the following : if three nondecreasing vectors are mutually orthogonal, one of them is the zero vector.

This question is the “discrete” version of that recent question.

What I have achieved so far : I can show the result when one of the vectors has sum $0$. Indeed, the following stronger result holds in this case :

Theorem If two nondecreasing vectors $u$ and $v$ are orthogonal and the “integral” (sum of coordinates) of $u$ is zero, then either $u$ is the zero vector or $v$ is “constant” (all its coordinates are equal).

Proof Suppose that $u=(u_1,u_2, \ldots ,u_n)$ is orthogonal to $v=(v_1,v_2, \ldots, v_n)$ and that the sum of $u$ is $0$ but $u$ is not the zero vector. The set $\Omega=\lbrace i | u_i \lt 0\rbrace$ is nonempty, call $r$ its largest element. Then we can decompose $u=(-g_r,-g_{r-1}, \ldots ,-g_2,-g_1,0, \ldots, 0, h_1,h_2, \ldots ,h_s)$ where $s \leq n-r$ and $0 \lt g_1 \leq g_2 \leq \ldots \leq g_r$ and $0 \lt h_1 \leq h_2 \leq \ldots \leq h_s$, and $\sum_{i=1}^r g_i=\sum_{j=1}^r h_j$ (since the integral of $u$ is zero) ; call $S$ this common value. Then, the orthogonality of $u$ and $v$ implies that

$$ Sv_r=\sum_{i=1}^r g_i v_r \geq \sum_{i=1}^r g_i v_i = \sum_{j=1}^s h_j v_{n+1-s+j} \geq \sum_{j=1}^s h_j v_{n+1-s}=Sv_{n+1-s} $$

Now $v_r \leq v_{n+1-s}$ because $v$ is nondecreasing, so all the inequalities above must be equalities and $v$ is indeed constant.

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This question has been cross-posted at MO, and promptly answered there

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