This is just a comment, not an answer. (I don't have enough reputation to write comments.)
Some basic statements that you probably knew already:
If $P(X \leq 0) = 1$, then there will always be such $\alpha$. Likewise if $P(X \geq 0) = 1$, there will never be such $\alpha$. This is from concavity/convexity of $u(x)$.
If the law of X is symmetric about zero, there will always be such $\alpha$. In particular $\alpha = 1/2$ will be better than $\alpha = 0$ or 1.
I agree with Harald that a general answer may be too much to hope for. Are you more interested in sufficient conditions or necessary conditions? Is there a particular family of distributions that you care about? Do you expect $\beta$ to be very close to 1, or much larger?
If you're looking for a sufficient condition, maybe you could let
$f(\alpha) = E[u(\alpha X + (1-\alpha) Y)]$
and look for situations in which $f'(0) > 0$.