expected value of the dot product of normalized random vector and its mean Suppose $U$ is a random vector satisfying $\mathbb E[U] = \mu$ and $\mathrm{var}(\|U\|_2) \le V$. Let $\bar{U} = U / \|U\|_2$ and $\bar\mu = \mu / \|\mu\|$. What is a lower bound on $\mathbb E[\bar U^\intercal\bar\mu]$ in terms of $V$? 
(For what it's worth, I have in mind a setting where $V$ is quite small compared to $\|\mu\|_2$---seems like there should be some non-trivial bound in this case.)
 A: The best possible general bound is the trivial one, $-1$.  (This deserves to be called "trivial" because $\bar U^\prime \bar \mu \ge -1$ in any event.)  To demonstrate this, I offer examples whose bounds come arbitrarily close to $-1$.
The intuition is this: the mean of the normalized vector $\bar U$ can be the opposite of the mean of $U$ itself because the normalization can greatly expand values close to zero.  If we pick a vector $U$ that is far from zero in one direction a small fraction of the time and otherwise is very close to zero but in the opposite direction the rest of the time, then (a) the mean of $U$ can lie in this privileged direction but (b) the mean of $\bar U$ can be arbitrarily close to a unit vector in the opposite direction.  This will cause $\bar U^\prime \bar \mu$ generally to be close to $-1$.
This construction works in any number of dimensions.  I will provide an explicit example in one dimension.

Let $0\lt\epsilon \ll 1/3$.  This makes $p=2\epsilon/(1+\epsilon)$ a probability because $0 \lt p \lt 1$.  Let $U$ take on the value $1$ with probability $p$ and the value $-\epsilon$ with probability $1-p$.  Straightforward computations yield
$$\mu=\mathbb{E}(U) = \epsilon \gt 0,$$
$$\bar\mu = 1,$$
and 
$$V = \operatorname{Var}(||U||_2) = \frac{2\epsilon(1-3\epsilon+3\epsilon^2-\epsilon^3)}{(1+\epsilon)^2} \approx 2\mu.$$
The random variable $\bar U$ takes on the value $1$ with probability $p$ and $-1$ with probability $1-p$.  Therefore
$$\mathbb{E}(\bar U^\prime \bar\mu)=\mathbb{E}(\bar U)=2p-1 = -1 + \frac{4\epsilon}{1+\epsilon} \lt 0.$$
This value remains unchanged even when we scale $U$ by $\sigma \gt 0$.  That scaling changes $\mu$ to $\mu\sigma$ and $V$ to $V\sigma^2$, so by taking $\sigma^2$ small we may make $V\sigma^2$ as small as we like compared to $\mu\sigma$.  Moreover, the greatest lower bound of $\mathbb{\bar U^\prime \bar \mu}$ is $-1$, QED.
