Problem: If $a,b,c > 0$ prove that
\[
  \frac{1}{a^3+b^3+abc}
  + \frac{1}{b^3+c^3+abc}
  + \frac{1}{c^3+a^3+abc}
  \le \frac{1}{abc}.
\]

Solution: From $a^3 + b^3 \ge ab(a+b)$, the left-hand side becomes
\[
  \sum_{\text{cyc}} \frac{1}{a^3+b^3+abc}
  \le \sum_{\text{cyc}} \frac{1}{ab(a+b+c)}
  = \frac{1}{abc} \sum_{\text{cyc}} \frac{c}{a+b+c}
  = \frac{1}{abc}.
\]