Problem: Let $a$, $b$, $c$ be real numbers greater than or equal to $1$.
Prove that
\[ \min \left( \frac{10a^2-5a+1}{b^2-5b+10},
  \frac{10b^2-5b+1}{c^2-5c+10},
  \frac{10c^2-5c+1}{a^2-5a+10} \right) \le abc. \]

Solution: Notice that
\[ \frac{10a^2-5a+1}{a^2-5a+10} \le a^3 \]
since it rearranges to $(a-1)^5 \ge 0$.
Cyclically multiply to get
\[
  \prod_{\text{cyc}} \left( \frac{10a^2-5a+1}{b^2-5b+10} \right)
  \le (abc)^3
\]
and the minimum is at most the geometric mean.