Berkeley Math Circle

Solution:
If we multiply the first equation by $(a b c)^{19}$ then it can be rewritten as
$$\left(a^{19} - b^{19}\right)\left(b^{19} - c^{19}\right)\left(c^{19} - a^{19}\right) \leq 0.$$
Similarly, the desired equation is equivalent to
$$\left(a^{20} - b^{20}\right)\left(b^{20} - c^{20}\right)\left(c^{20} - a^{20}\right) \leq 0.$$
These are evidently equivalent since $a^{19} - b^{19}$ and $a^{20} - b^{20}$ have the same sign, etc.