Solve an inequality $|x| + |x-y| \geq |x_0| + |x_0-y| $ for $y$

Question

Fix $x_0 \in \mathbb R^n$. Describe $y \in \mathbb R^n$ that satisfies $|x| + |x-y| \geq |x_0| + |x_0-y|$ for all $x \in \mathbb R^n$.

I have tried triangle inequalities (including reverse) repeatedly, but does not seem to work. How can I describe such $y$?

Answer 1

If $x_0=0$, clearly the inequality is satisfied by every vector $y$. Suppose $x_0\ne0$. If the given condition holds, in particular it must hold for $x=0$. Hence \begin{align*} |y| = |0| + |0-y| \geq |x_0| + |x_0-y|\ge|y|. \end{align*} Therefore, equality holds in the triangle inequality $|x_0|+|y-x_0|\ge|y|$, meaning that $y-x_0$ is a nonnegative multiple of $x_0$. So we conclude that $y=kx_0$ for some $k\ge1$.

It remains to verify that the original inequality indeed holds when $y=kx_0$ for some $k\ge1$. I will leave it to you.

Answer 2

Geometric consideration shows that $y$ and $x_0$ are co-linear and for this purpose, analytically, let $x=\lambda y+(1-\lambda)x_0$ for $\lambda>0$ then $$|\lambda y+(1-\lambda)x_0| + |\lambda y+(1-\lambda)x_0-y| \geq |x_0| + |x_0-y|$$ $$|\lambda y+(1-\lambda)x_0| + (1-\lambda)|x_0-y| \geq |x_0| + |x_0-y|$$ $$|x_0-\lambda(x_0-y)| \geq |x_0| + \lambda|x_0-y|$$ this inequality valid only if $\lambda(x_0-y)=k x_0$, for a real $k$.