Suppose w.l.o.g. that $z \ge x$ and $z \ge y$. From the second equation $\frac{1}{x} \ge 2013$,

therefore $x \le \frac{1}{2013}$. From the third equation $\frac{1}{y} \le 2013$, due to which $y \ge \frac{1}{2013} \ge x$.
But now from the first equation $\frac{1}{z} \ge 2013$, therefore $z \le \frac{1}{2013}$. Since we assumed that $z \ge y \ge \frac{1}{2013}$, the only possibility is $z = y = \frac{1}{2013}$, then also $x = \frac{1}{2013}$.

Adding up all the equations, we get $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3 \cdot 2013$. Multiplying the first equation by $z$, the second one by $x$ and third one by $y$ and adding together we get $3 = 2013(x + y + z)$. Therefore the arithmetic and harmonic mean of $x,y,z$ are both equal to $\frac{1}{2013}$. Consequently $x = y = z = \frac{1}{2013}$.