I encountered an inequality with the following variables and functions. $X$ is a random variable drawn from $(-\infty, \infty)$ with cdf $F$, pdf $f$, and mean $\mu=\mathbb{E}[X]$. For any $x$ and $\alpha \in[0,\pi/2]$, $z(x)$ is a solution to \begin{equation*} z(x) f\left(u\right) - \left[1 - F\left(u\right) \right] \cos \alpha=0, \end{equation*} where $u = \frac{z(x)- x \sin \alpha - \mu (1-\sin \alpha - \cos \alpha)}{\cos \alpha}.$
Then, for any $\alpha\in(0,\pi/2)$ show that \begin{equation*} \int_{-\infty}^{\infty} \left[1-F\left(u\right) \right] \Big(x - z(x) \sin \alpha - \mu (1-\sin \alpha)\Big)dF(x)>0. \end{equation*}
Attempt: Rewrite terms inside the large curly brackets as $(x-\mu)(1-\sin \alpha) + (x-z(x)) \sin \alpha$ and, hence, rewrite the integral as a sum of two integrals, each containing one of the above terms. It is easy to show that the integral term containing $(x-\mu)(1-\sin \alpha)$ is positive, but I could not show that the intergral term containing $(x-z(x)) \sin \alpha$ is also positive.