Is the following inequality true or not?

Question

$\left\|\sum_{m=1}^{M}\left(x_{m}^{t+1}-x_{m}^{t}\right)\right\|^{2}$ $\leq \sum_{m=1}^{M}\left\|x_{m}^{t+1}-x_{m}^{t}\right\|^{2}$, where $\left\|\cdot\right\|$ denote the 2-norm, and the $x_{m}^{t}$ is the vector.

Answer 1

The inequality is false for the same reason that $(1+2)^{2} \leq 1^{2}+2^{2}$ is false.

Answer 2

Apply the Cauchy-Schwartz inequality:

$$\left(\sum_{i=1}^{n} x_{i} y_{i}\right)^{2} \leq\left(\sum_{i=1}^{n} x_{i}^{2}\right)\left(\sum_{i=1}^{n} y_{i}^{2}\right)$$

The left hand of our inequality can be represented as follows.

$ \left\|\sum_{m=1}^{M}\left(x_{m}^{t+1}-x_{m}^{t}\right)\right\|^{2} = \left\|\sum_{m=1}^{M}\left(x_{m}^{t+1}-x_{m}^{t}\right)\times 1\right\|^{2} \\ \leq \sum_{m=1}^{M}\left\|x_{m}^{t+1}-x_{m}^{t}\right\|^{2} \times \sum_{m=1}^{M}1^2= M\sum_{m=1}^{M}\left\|x_{m}^{t+1}-x_{m}^{t}\right\|^{2} $

Therefore, the origin inequality lacks M.