If $\psi(x)=\int_U d(x,y)dy$, proof that there is a unique $x\in \Bbb R$ such that $\psi(x)$ is minimal?

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Let $(\Bbb R,d)$ be a metric space and let $\psi:\Bbb R\to[0,\infty)$ be defined as $\psi(x)=\int_Ud(x,y)dy$ given a subset $U\subseteq \Bbb R$ that is a bounded member of the usual topology on $\Bbb R$.

I'm trying to prove that there is a unique $x\in\Bbb R$ such that $\psi(x)$ is minimum, i.e. for all other $y\in\Bbb R$, we have that $\psi(x)<\psi(y)$, and not just $\psi(x)\leq\psi(y)$.

If $\psi$ were injective then we are done, but it's not always the case. I need to check injectiveness just for the argument that just gives the minimal, but I don't know to do that. Any help on that or any other methods would be appreciated. Thanks.