Use Titu's lemma and then just prove that $(a^{1007}+b^{1007}+c^{1007})\geq 3$ (AM $\geq$ GM and $abc=1$)

$$\sum_{cyc}\frac{a^{2014}}{1+2bc}\overset{\text{Generalized T2'S}}{\ge}\frac{\left(\sum_{cyc}a\right)^{2014}}{3^{2012}\left(3+2\sum_{cyc}ab\right)}\text{ furthermore the inequality becomes }\left(\sum_{cyc}a\right)^{2014}\sum_{cyc}ab\ge3^{2013}\left(3+2\sum_{cyc}ab\right)$$$$\text{ Now using the fact that }abc=1\text{ we obtain }\left(\sum_{cyc}a\right)^{2014}\sum_{cyc}ab\ge3^{2014}\sum_{cyc}ab$$$$\text{ Thus the inequality transforms into }3\sum_{cyc}ab\ge3+2\sum_{cyc}ab\Longrightarrow\sum_{cyc}ab\ge3\text{ which is just AM-GM }$$