Requested in comments:
Clearly in general $\mathbb{E}[Z\mid Z\le \tau] \le \tau \lt \mathbb{E}[Z\mid Z>\tau]$, assuming the expectations exist.
If $\tau$ is the median of $Z$ then:
- $\mathbb P(Z>\tau) \le \frac12 \le \mathbb P(Z\le \tau)$
- $\mathbb{E}[Z] = \mathbb P(Z>\tau)\,\mathbb{E}[Z\mid Z>\tau] + \mathbb P(Z\le\tau)\,\mathbb{E}[Z\mid Z\le\tau]$
- so $\mathbb{E}[Z] \le \frac12 \mathbb{E}[Z\mid Z>\tau] + \frac12 \mathbb{E}[Z\mid Z\le\tau] $
- and thus $\mathbb{E}[Z] \le \frac12 \mathbb{E}[Z\mid Z>\tau]+ \frac12 \tau $
- implying $\mathbb{E}[Z\mid Z>\tau] \ge 2 \mathbb{E}[Z] -\tau$
- which combined with the earlier result gives $$\mathbb{E}(Z\mid Z>\tau)\ge \max(\tau,2\mathbb{E}(Z) -\tau).$$
If you extended this to the $p$th quantile of $Z$ so $\mathbb P(Z\le \tau) \ge p$ and $\mathbb P(Z>\tau)\le 1-p$ then the same argument would give you $\mathbb{E}[Z\mid Z>\tau] \ge \max\left(\tau,\frac{\mathbb{E}[Z]-p\tau}{1-p}\right). $