Let $Z$ be a symmetric random variable with density $\pi$ where $\pi(t) > p_{min}$ for all $t \in [-\tau, \tau]$ for $p_{min}, \tau > 0$. Define $\phi(t) = \mathbb E [|t+Z| - |Z|]$.
Show that $\phi(t) \geq 2 p_{min}h_\tau(t)$ where $h_\tau(t)$ is the Huber loss:
$h_\tau(t) = 1/2t^2$ if $|t| \leq \tau$ and $\tau|t| - 1/2\tau^2$ otherwise.
What I've Tried
Assuming that the density of $Z$ is absolutely continuous with respect to Lebesgue measure, we can apply the mean value theorem to $\phi$ since it is differentiable almost everywhere:
$\phi(t) = \mathbb E [|t+Z| - |Z|] = \mathbb E [\nabla |t|_{t = Z}t]$ where the second derivative term is $0$ since $|t|$ has curvature of $0$ almost everywhere.
But $\mathbb E [\nabla |t|_{t = Z}t] = t(\mathbb P(Z>0) - \mathbb P(Z<0)) = 0$ since $Z$ is symmetric. I've now shown that $\phi(t) = 0$ for all $t$.
Where did I go wrong?