Assume that $x$ is the minimum among $x$, $y$ and $z$. 1) $x=3$: Because $3^5+1=244=2^2\times61$, we would know that $z=61$. Also we know that $3|(y^5+1)$, so $y\equiv-1(mod\ 3)$.Obviously, $5\nmid(61^5+1)$, and according to calculation we know that $11|(61^5+1)$. $y_{min}=11$, $(xyz)_{min}=3\times11\times61=2013$. Which means that the minimum set of three consecutive primes $x,\ y,\ z$ saisfying $xyz>2013$ should be $11,\ 13,\ 17$. Therefore, we only need to talk about when $x=5$ and $x=7$. 2) $x=5$: $5^5+1=3$, $126=2\times3\times521$, but $5\times521>2013$, which does not satisfy our needs. 3) $x=7$: Because $7|(y^5+1)$, we know that $y\equiv-1(mod\ 7)$. If $y=13$, then $13|(z^5+1)$. So, $z\equiv-1(mod\ 13)$. Therefore, $z_{min}=103$, $(xyz)_{min}>2013$. According to above, the minimum value of $xyz$ is 2013.