Easy way to see that $|x-z|^2 \geq |y-z|^2 + (2y-2z)\cdot (x-y),\forall x,y \in \mathbb{R}^n$, for fixed $z \in \mathbb{R}^n$

Question

I am having a hard time seeing this inequality. I have tried completing the square, or finding some way to fit in the triangle inequality. It seems like I am missing something that is apparently supposed to be clear.

Answer 1

Hint: For $x,y,z\in\mathbb{R}^n$, we have $$\|x-z\|^2=\big\|(y-z)+(x-y)\big\|^2=\|y-z\|^2+2\,(y-z)\cdot(x-y)+\|x-y\|^2\,.$$ Here, I assume that $\|\_\|$ (or $|\_|$ in your notation) is the usual Euclidean norm of $\mathbb{R}^n$.

Answer 2

Let $x(x_1,x_2,...,x_n)$, $y(y_1,y_2,...,y_n)$ and $z(z_1,z_2,...z_n)$.

Thus, we need to prove that $$\sum_{i=1}^n(x_i-y_i)(x_i+y_i-2z_i)\geq2\sum_{i=1}^n(x_i-y_i)(y_i-z_i)$$ or $$\sum_{i=1}^n(x_i-y_i)^2\geq0.$$