How an $\ell_1$ Inequality Implies Equality

How an $\ell_1$ Inequality Implies Equality

Suppose that for scalar $\epsilon$ we know that $\vert \epsilon \vert$ is
small enough such that the sign pattern on $\mathbf{x}\in\mathbb{R}^n$ is
equal to that on $\mathbf{x} + \epsilon \mathbf{h}$, which makes the
$\ell_1$ norm continuous and differentiable in this region. Now, if it is
true that
$$ \Vert \mathbf{x} \Vert_1 \leq \Vert \mathbf{x} + \epsilon \mathbf{h}
\Vert_1 $$
for each such $\epsilon$, must it be true that $\Vert \mathbf{x} \Vert_1 =
\Vert \mathbf{x} + \epsilon \mathbf{h} \Vert_1$?
What I am reading claims that it is and uses the following wording:
"[Since] the above relationship [referring to the inequality] holds for
both positive and negative values of $\epsilon$ in a region where the
$\ell_1$ function is continuous and differntiable, [...] the only way this
could be true is if the above inequality is satisfied as an equality."
I don't understand how this conclusion is drawn. If $f$ is continuous and
differentiable, just because $f(x)\leq f(x \pm \delta)$ does not mean that
$f(x)=f(x\pm \delta)$.