Berkeley Math Circle Monthly Contest 1

Solution:

Using $a b c = 1$, we transform the given condition to
$$a^{n} + b^{n} + c^{n} > \frac{1}{a^{n}} + \frac{1}{b^{n}} + \frac{1}{c^{n}} .$$
$$a + b + c > b c + c a + a b$$
or
$$- b c - c a - a b + a + b + c > 0 .$$
We then add $a b c - 1 (= 0)$ to the left side, getting
$$a b c - b c - c a - a b + a + b + c - 1 > 0$$
which factors as
$$(a - 1)(b - 1)(c - 1) > 0 .$$
In exactly the same way we transform the condition to be proved to
$$\left(a^{n} - 1\right)\left(b^{n} - 1\right)\left(c^{n} - 1\right) > 0 .$$
However, for any positive real $x$, the numbers $x - 1$ and $x^{n} - 1$ are both positive, both negative, or both zero. Consequently (1) and (2) are equivalent.