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R.P.
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\operatorname{} has context-dependent spacing an in some other respects does not always have the same effect as \mbox{}. Also some other minor edits.
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For any given $x \in \mathbb{R}^n$, let $\nabla{x} \in \mathbb{R}^{n \choose 2}$ be the vector whose {i,j}$\{i,j\}$-th entry is $|x_i-x_j|$. I think the following claim is true.

Claim. If $f, g \in \mathbb{R}^n$ are vectors with zero mean, i.e., $$\sum_{i=1}^n f_i = \sum_{i=1}^n{g_i}=0$$$$\sum_{i=1}^n f_i = \sum_{i=1}^n g_i = 0$$ and the angle between them is at most $\frac{\pi}{2}$, then $$\mbox{dist} (\nabla{f},\nabla{g}) \ge \mbox{dist}(f,g).$$$$\operatorname{dist} (\nabla{f},\nabla{g}) \ge \operatorname{dist}(f,g).$$

If anybody has any idea about how to prove this, please share it with me. Thanks.

For any given $x \in \mathbb{R}^n$, let $\nabla{x} \in \mathbb{R}^{n \choose 2}$ be the vector whose {i,j}-th entry is $|x_i-x_j|$. I think the following claim is true.

Claim. If $f, g \in \mathbb{R}^n$ are vectors with zero mean, i.e., $$\sum_{i=1}^n f_i = \sum_{i=1}^n{g_i}=0$$ and the angle between them is at most $\frac{\pi}{2}$, then $$\mbox{dist} (\nabla{f},\nabla{g}) \ge \mbox{dist}(f,g).$$

If anybody has any idea about how to prove this, please share it with me. Thanks.

For any given $x \in \mathbb{R}^n$, let $\nabla{x} \in \mathbb{R}^{n \choose 2}$ be the vector whose $\{i,j\}$-th entry is $|x_i-x_j|$. I think the following claim is true.

Claim. If $f, g \in \mathbb{R}^n$ are vectors with zero mean, i.e., $$\sum_{i=1}^n f_i = \sum_{i=1}^n g_i = 0$$ and the angle between them is at most $\frac{\pi}{2}$, then $$\operatorname{dist} (\nabla{f},\nabla{g}) \ge \operatorname{dist}(f,g).$$

If anybody has any idea about how to prove this, please share it with me. Thanks.

Notice added Canonical answer required by j.s.
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Improved the appearance of {i,j}
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Taras Banakh
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For any given $x \in \mathbb{R}^n$, let $\nabla{x} \in \mathbb{R}^{n \choose 2}$ be the vector whose ${\{i,j\}}${i,j}-th entry is $|x_i-x_j|$. I think the following claim is true.

Claim. If $f, g \in \mathbb{R}^n$ are vectors with zero mean, i.e., $$\sum_{i=1}^n f_i = \sum_{i=1}^n{g_i}=0$$ and the angle between them is at most $\frac{\pi}{2}$, then $$\mbox{dist} (\nabla{f},\nabla{g}) \ge \mbox{dist}(f,g)$$$$\mbox{dist} (\nabla{f},\nabla{g}) \ge \mbox{dist}(f,g).$$

If anybody has any idea about how to prove this, please share it with me. Thanks.

For any given $x \in \mathbb{R}^n$, let $\nabla{x} \in \mathbb{R}^{n \choose 2}$ be the vector whose ${\{i,j\}}$th entry is $|x_i-x_j|$. I think the following claim is true.

Claim. If $f, g \in \mathbb{R}^n$ are vectors with zero mean, i.e., $$\sum_{i=1}^n f_i = \sum_{i=1}^n{g_i}=0$$ and the angle between them is at most $\frac{\pi}{2}$, then $$\mbox{dist} (\nabla{f},\nabla{g}) \ge \mbox{dist}(f,g)$$

If anybody has any idea about how to prove this, please share it with me. Thanks.

For any given $x \in \mathbb{R}^n$, let $\nabla{x} \in \mathbb{R}^{n \choose 2}$ be the vector whose {i,j}-th entry is $|x_i-x_j|$. I think the following claim is true.

Claim. If $f, g \in \mathbb{R}^n$ are vectors with zero mean, i.e., $$\sum_{i=1}^n f_i = \sum_{i=1}^n{g_i}=0$$ and the angle between them is at most $\frac{\pi}{2}$, then $$\mbox{dist} (\nabla{f},\nabla{g}) \ge \mbox{dist}(f,g).$$

If anybody has any idea about how to prove this, please share it with me. Thanks.

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Tom Goodwillie
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