This is a cross-posting of a MSE [question](https://math.stackexchange.com/questions/311376/orthogonality-between-three-vectors-whose-coordinates-are-nondecreasing) (which did not receive any feedback there so far).


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.



Orthogonality between vectors whose components increase
Consider the nondecreasing vector $v=(1,1,\ldots,1)$. It's not too hard to show that two nondecreasing nonzero vectors orthogonal to $v$ must have a positive inner product. Here is a proof.


Let $a=(a\_1,\ldots, a\_n)$ and $b=(b\_1,\ldots, b\_n)$ be orthogonal to $v$. Suppose that $a\_j$ is the last negative entry of $a$ and $b\_k$ is the last negative entry of $b$, for $k>j$. Then


\begin{equation}\sum\_{i=1}^j |a\_ib\_i| \ge \sum\_{i=1}^j |a\_i b\_{j+1}| \ge \sum\_{i=j+1}^k |a\_ib\_{j+1}|\ge \sum\_{i=j+1}^k |a\_ib\_i|\end{equation}


But the sum defining the inner product is


\begin{equation}\sum\_{i=1}^j |a\_ib\_i| - \sum\_{i=j+1}^k |a\_ib\_i| + \sum\_{i=k+1}^n |a\_ib\_i|\end{equation}


So this is nonzero as long as $a\_n$ and $b\_n$ are nonzero; since $a$ and $b$ are orthogonal to $v$, $a\_n$ is zero if and only if $a$ is zero, and likewise for $b.$


Now, a generic nondecreasing vector is of the form $av+u$ for $u$ nondecreasing and orthogonal to $v$.


Assume we have three mutually orthogonal nonzero nondecreasing vectors $v\_i= a\_i v +u\_i$ for $i\in\{1,2,3\}$. No more than one $u\_i$ can be zero; suppose $u\_1$ and $u\_2$ are nonzero. Then $a\_1$ and $a\_2$ must have opposite signs for $v\_1$ and $v\_2$ to be orthogonal (in particular, neither can be zero). Now $u\_3$ cannot be zero because then $v\_3$ would not be orthogonal to $v\_1$ or $v\_2$. Then by the same argument, $a\_3$ must have a sign opposite to both $a\_1$ and $a\_2$, a contradiction.


