It is equivelant $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1+ \frac{2015}{xyz}$. Let $x \ge y \ge z$. WLOG $z \ge 3 \Rightarrow 1 < 1+ \frac{2015}{xyz} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \le 1$. Contradiction. For $z=2$ we have $0 \le (x-2)(y-2)= -2011$. Contradiction. For $z=1$ we have $x+y= 2015$. Conclusion follows