Let H an infinite-dimensional Hilbert space in $\mathbb{R}$

If $x_1, x_2, \ldots x_n \in H$, how to prove:

$\sum_{1\leq i,j\leq n} {\lvert\lvert x_i - x_j \rvert\rvert}^2 \leq \sum_{1\leq i,j\leq n} ({\lvert\lvert x_i \rvert\rvert}^2 + {\lvert\lvert x_j \rvert\rvert}^2)$

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    $\begingroup$ I'm completely blanking on how this isn't just the triangle inequality. $\endgroup$ – Titus Jan 14 '16 at 1:59

We have \begin{align*}\sum_{1\le i,j\le n} \|x_i - x_j\|^2 &=\sum_{1\le i,j\le n} \left\{\|x_i\|^2 + \|x_j\|^2 - 2 \, (x_i,x_j)\right\} \\&= \sum_{1\le i,j\le n} \left\{\|x_i\|^2 + \|x_j\|^2\right\} - 2 \, (\sum_{i=1}^n x_i,\sum_{j=1}^n x_j) \\&\le \sum_{1\le i,j\le n} \left\{\|x_i\|^2 + \|x_j\|^2\right\}\end{align*}


The triangle inequality yields only $\|x_i - x_j\|^2 \le (\|x_i\| + \|x_j\|)^2 \le \|x_i\|^2 + 2 \|x_i\| \|x_j\| + \|x_j\|^2 \le 2 (\|x_i\|^2 + \|x_j\|^2)$...

  • $\begingroup$ $(|x|+|y|)^2$ is not equal to $|x|^2+2\langle x,y \rangle+|y|^2$. $\endgroup$ – wj32 Jan 17 '16 at 11:30
  • $\begingroup$ Of course not... Corrected now. $\endgroup$ – mkk030572 Jan 17 '16 at 11:32

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