Problem: Let $ABC$ be a triangle.
Its excircles touch sides $BC$, $CA$, $AB$ at $D$, $E$, $F$.
Prove that the perimeter of triangle $ABC$ is
at most twice that of triangle $DEF$.

Solution: Solution by August Chen:
It turns out that it is enough to take the
orthogonal projection of $EF$ onto side $BC$
(which has length $a-(s-a)(\cos B + \cos C)$)
and sum cyclically:
\begin{align*}
  -s + \sum_{\text{cyc}} EF &\ge
  -s +
  \sum_{\text{cyc}}
  \left[ a - (s-a)\left( \cos B + \cos C \right) \right] \\
  &= s - \sum_{\text{cyc}} a \cos A
    = \sum_{\text{cyc}} a \left( \half - \cos A \right) \\
  &= R \sum_{\text{cyc}}
  \sin A (1 - 2 \cos A) \\
  &= R \sum_{\text{cyc}} \left( \sin A - \sin2A \right).
\end{align*}
Thus we're done upon noting that
\[ \frac{\sin2B + \sin2C}{2} = \sin(B+C) \cos(B-C)
  = \sin A \cos(B-C) \le \sin A. \]
(Alternatively, one can avoid trigonometry by substituting
$\cos A = \frac{b^2+c^2-a^2}{2bc}$
and doing some routine but long calculation.)