Is it possible that $\forall y \in X \quad \|y-b\| < \max(\|y-a\|, \|y+a\|) $?

Question

Let $(X, \|\cdot\|)$ be a normed space. Let $a, b \in X$ be noncollinear vectors. Is it possible that \begin{equation} \forall y \in X \quad \|y-b\| < \max(\|y-a\|, \|y+a\|)? \end{equation}

So far, I wasn't able to find a norm which would satisfy that and I know that this isn't true in normed spaces with norm generated by inner product - exemplary vector $y \in X$ for which the statement is false can be found on line given by the equation $y = b + \beta x$, where $\langle\beta, a\rangle = 0$ and $\beta \ne 0$.

Answer 1

What about $X=\mathbb{R}^2$, but with $||(x_1,x_2)||=\max(|x_1|,|x_2|)$?

Consider $a=(2,2)$ and $b=(-1,1)$. Then $||y-b||=\max(|y_1+1|,|y_2-1|)$.

Clearly $|y_1+1|<\max(|y_1-2|,|y_1+2|)$ and $|y_2-1|<\max(|y_2-2|,|y_2+2|)$.

Since $\max(||y-a||,||y+a||)=\max(|y_1-2|,|y_2-2|,|y_1+2|,|y_2+2|)$ we conclude the desired result.

The "cheat" here is that on both coordinates of $\mathbb{R}^2$, $a$ and $b$ are collinear.