When $xy + yz + zx = 1$ and $(x + y)(y + z)(z + x) \neq 0$, we have
$$\frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} \ge \frac{5}{2}. \qquad (1)$$
Now, WLOG we can assume $a^{2016}b^{2016}+b^{2016}c^{2016}+c^{2016}a^{2016} = 1$. Then, we have
$$\frac{1}{a^{2016} + b^{2016}} + \frac{1}{b^{2016} + c^{2016}} + \frac{1}{c^{2016} + a^{2016}} \ge \frac{5}{2}. \qquad (2)$$
Furthermore, by Muirhead and Schur's inequality of degree 3, we have
$$(a+b+c)^{4032} \\ \ge (a^{4032} + b^{4032} + c^{4032} + 3a^{1344} + b^{1344} + c^{1344}) + \sum_{i=1}^{2015} C_i^{2016} (a^i b^{2016-i} + b^i c^{2016-i} + c^i a^{2016-i}) \\ \ge 2^{4032} (a^{2016}b^{2016} + b^{2016}c^{2016} + c^{2016}a^{2016}) = 2^{4032}.$$
Combining these, we get the minimum value is $5 \times 2^{2015}$.