I would like to get some help with the next problem:
I'm trying to prove that $$\sum_{i = 1}^n (x_i - y_i)^2 \le \sum_{i = 1}^n (x_i - z_i)^2 + \sum_{i = 1}^n (z_i - y_i)^2\;\;\;\;\;(1).$$
I tried this: $$ \sum_{i = 1}^n (x_i - y_i)^2 =$$ $$= \sum_{i = 1}^n [x_i^2 -2x_iy_i + y_i^2 + (-2x_iz_i + z_i^2) + (z_i^2 - 2z_iy_i) - (-2x_iz_i + z_i^2) - (z_i^2 - 2z_iy_i)] =$$ $$= \sum_{i = 1}^n [(x_i^2 - 2x_iz_i + z_i^2) + (z_i^2 - 2z_iy_i + y_i^2) - 2z_i^2 -2x_iy_i + 2x_iz_i + 2z_iy_i] =$$ $$= \sum_{i = 1}^n (x_i - z_i)^2 + \sum_{i = 1}^n (z_i - y_i)^2 - 2\sum_{i = 1}^n z_i^2 - 2\sum_{i = 1}^n (x_iy_i - x_iz_i - z_iy_i).$$
I wanted to get an expression from which it would be easier to conclude that $\sum_{i = 1}^n (x_i - y_i)^2$ is smaller than the right side of the inequality $(1)$, but the last expression isn't useful.
Please, could you give me some advice about what should i do to prove this inequality?
EDIT: I apologise, because i made a huge mistake. It wasn't explicitly stated, so i oversaw it: I need to prove a different inequality: $$\sqrt{\sum_{i = 1}^n (x_i - y_i)^2} \le \sqrt{\sum_{i = 1}^n (x_i - z_i)^2} + \sqrt{\sum_{i = 1}^n (z_i - y_i)^2}.$$
EDIT 2: I think i solved it. Using Minkowski inequality, i got: $$\sqrt{\sum_{i = 1}^n (x_i - y_i)^2} = \sqrt{\sum_{i = 1}^n |x_i + (-y_i)|^2} = \sqrt{\sum_{i = 1}^n |x_i - z_i + z_i + (-y_i)|^2} \le \sqrt{\sum_{i = 1}^n |x_i - z_i|^2} + \sqrt{\sum_{i = 1}^n |z_i - y_i|^2} = \sqrt{\sum_{i = 1}^n (x_i - z_i)^2} + \sqrt{\sum_{i = 1}^n (z_i - y_i)^2}.$$
It is enought to prove for $n=1$: $$(x-y)^2\leq (x-z)^2+(z-y)^2$$
It is the same as $$0\leq 2z^2-2xz-2yz+2xy$$
and this is $$(z-x)(z-y)\geq 0$$ which is not true if, say $x=1,z=2$ and $y=3$.
So you have forgotten something.