Let $X$ be a random variable indicating the gains in some investment, let $\mu = E[X]$ be the expected value, and $U(X)$ is the utility obtained from that gain.
If $u$ is convex, then $U^{\prime \prime} \geq 0$, and from Jensen's inequality, we have $E[U(X)] \geq U(E[X])$. Thus, it is better to take $X$ itself rather than to get $\mu = E[X]$ for sure.
However, if we were allowed to get an arbitrary mixture of $X$ and $\mu$, with expected utility $E[U(\alpha X + (1-\alpha) \mu)]$, where $0 \leq \alpha \leq 1$, then is it possible that this expected utility is maximized at some $\alpha$ strictly less than 1? I suspect the answer is "no", but I'm not sure how to prove it.
(This is problem 8.22 in "First course in probability" by Sheldon Ross.)
let $f(\alpha) \equiv E[U(\alpha X + (1-\alpha) \mu)]$. Differentiating, we have $f^{\prime \prime}(\alpha) = E[(X - \mu)^2 U^{\prime \prime}(\alpha X + (1-\alpha) \mu)]$ and thus $f^{\prime \prime}$ always has the same sign as $U^{\prime \prime}$.
Hence, if $U$ is convex, then $f$ permits no local maxima at $0 < \alpha < 1$.