Why can we claim that $c^{2}\mathbb{V}(U) + 2c\mathbb{E}(TU) \geq 0$ implies that $\mathbb{E}(TU) = 0$?

Question

Suppose that $T$ is a UMVUE for $\theta$. Then let $T_{c} = T + cU$ where $U$ is an unbiased estimator of zero and $c\in\mathbb{R}$ is a fixed constant. Hence $T_{c}$ is also unbiased for $\theta$, whence we conclude that \begin{align*} \mathbb{V}(T_{c}) \geq \mathbb{V}(T), \end{align*} for all $c\in\mathbb{R}$ and every $\theta\in\Theta$ (the parametric space). According to the properties of variance, it is the same as to claim that \begin{align*} c^{2}\mathbb{V}(U) + 2c\text{Cov}(T,U) \geq 0, \end{align*} for every $c\in\mathbb{R}$ and $\theta\in\Theta$. However, since $\mathbb{E}(U) = 0$, we can conclude that \begin{align*} \begin{cases} \mathbb{V}(U) = \mathbb{E}(U^{2}) - [\mathbb{E}(U)]^{2} = \mathbb{E}(U^{2})\\\\ \text{Cov}(T,U) = \mathbb{E}(TU) - \mathbb{E}(T)\mathbb{E}(U) = \mathbb{E}(TU) \end{cases} \end{align*} Consquently, one arrives at the follow relation: \begin{align*} c^{2}\mathbb{E}(U^{2}) + 2c\mathbb{E}(TU) \geq 0 \end{align*} But then I get stuck. How do we conclude from here that $\mathbb{E}(TU) = 0$?

Answer 1

The last line is a quadratic in $c$, which is $\ge 0$ for all $c$ and $=0$ for some $c$ ($c=0$). So its discriminant must be zero.