For any three vectors $x,y,z\in\mathbb{R}^d$, we have $ \|y-z\|\cdot\|x\|\leq\|x-y\|\cdot\|z\|+\|z-x\|\cdot\|y\|$

Does anyone know a proof of the following problem?

Problem: Show that for any three vectors ${\bf x}, {\bf y}, {\bf z}\in \mathbb{R}^d$ the following holds, $$ \|{\bf y} - {\bf z}\|\cdot \|{\bf x}\| \leq \|{\bf x} - {\bf y}\|\cdot \|{\bf z}\| + \|{\bf z} - {\bf x}\|\cdot\|{\bf y}\|.$$

All of the norms are Euclidean 2-norm.

Solution 1:

If any of $x,y,$ or $z$ is zero the result is trivial. So WOLOG assume that $x,y,$ and $z$ are all non-zero and let $a = x/\Vert x\Vert^2$, $b=y/\Vert y\Vert^2$, and $c = z/\Vert z\Vert^2$. We have $\left\Vert x-y\right\Vert=\frac{\left\Vert a-b\right\Vert}{\Vert a\Vert\Vert b\Vert}$, $\left\Vert z-x\right\Vert=\frac{\left\Vert c-a\right\Vert}{\Vert c\Vert\Vert a\Vert}$, and $\left\Vert y-z\right\Vert=\frac{\left\Vert b-c\right\Vert}{\Vert b\Vert\Vert c\Vert}$. Therefore, $$\left\Vert y-z\right\Vert\Vert x\Vert=\frac{\left\Vert b-c\right\Vert}{\Vert a\Vert\Vert b\Vert\Vert c\Vert}\\\leq\frac{\left\Vert a-b\right\Vert}{\Vert a\Vert\Vert b\Vert\Vert c\Vert}+\frac{\left\Vert c-a\right\Vert}{\Vert a\Vert\Vert b\Vert\Vert c\Vert}\\=\left\Vert x-y\right\Vert\Vert z\Vert + \left\Vert z-x\right\Vert\Vert y\Vert,$$ which completes the proof.