Final Round of National Olympiad

Let $a^{2014} = u$, $b^{2014} = v$ and $c^{2014} = w$; then $abc = 1$ gives that $uvw = 1$. As the numerators of the l.h.s. of the inequality to be proven are positive, the inequality is equivalent to
$$(1+v)(1+w) + (1+w)(1+u) + (1+u)(1+v) > (1+u)(1+v)(1+w).$$

By expanding, simplifying and using $uvw = 1$, we get $1 + u + v + w > 0$. But this is true, since $u$, $v$ and $w$ are positive.

Let $a^{2014} = u$, $b^{2014} = v$ and $c^{2014} = w$. Then $abc = 1$ gives that $uvw = 1$. Therefore there exist positive real numbers $x$, $y$, $z$ such that $u = \frac{x}{y}$, $v = \frac{y}{z}$ and $w = \frac{z}{x}$.
The inequality can then be written as $\frac{1}{1+\frac{x}{y}} + \frac{1}{1+\frac{y}{z}} + \frac{1}{1+\frac{z}{x}} > 1$ which is equivalent to
$$\frac{y}{x+y} + \frac{z}{y+z} + \frac{x}{z+x} > 1.$$
But this inequality can be obtained by adding the obvious inequalities $\frac{y}{x+y} > \frac{y}{x+y+z}$, $\frac{z}{y+z} > \frac{z}{x+y+z}$ and $\frac{x}{z+x} > \frac{x}{x+y+z}$.

Let $a^{2014} = u$, $b^{2014} = v$ and $c^{2014} = w$; then $abc = 1$ gives that $uvw = 1$. W.l.o.g., let $w$ be the greatest of $u$, $v$, $w$. Then $w \ge 1$, because otherwise $u$, $v$, $w$ would all be less than $1$ and their product could not be $1$. Thus $uv \le 1$. Now
$$\frac{1}{1+u} + \frac{1}{1+v} - 1 = \frac{(1+v) + (1+u) - (1+u)(1+v)}{(1+u)(1+v)} = \frac{1-uv}{(1+u)(1+v)} \ge 0.$$
Thus $\frac{1}{1+u} + \frac{1}{1+v} \ge 1$, from which the desired inequality can be concluded.