Problem: Let $ABC$ be a triangle and $P$ any point
such that $PA$, $PB$, $PC$ are the sides of an obtuse triangle,
with $PA$ the longest side.
Prove that $\angle BAC$ is acute.

Solution: Using Ptolemy's inequality and Cauchy-Schwarz,
\begin{align*}
  PA \cdot BC
  &\le PB \cdot AC + PC \cdot AB \\
  &\le \sqrt{(PB^2+PC^2)(AB^2+AC^2)} \\
  &< \sqrt{PA^2 \cdot (AB^2+AC)^2} = PA \cdot \sqrt{AB^2+AC^2}
\end{align*}
meaning $BC^2 < AB^2+AC^2$, so $\angle BAC$ is acute.

\begin{remark*}
  [Lokman G\"{o}k\c{c}e]
  Here is another approach using Euler's quadrilateral formula.
  Let $M$ and $N$ be midpoints of $AP$ and $BC$, respectively.
  For the points $A$, $B$, $P$, $C$; apply Euler's quadrilateral formula to get
  \[ AB^2 + BP^2 + PC^2 + CA^2 = AP^2 + BC^2 + 4MN^2 \geq AP^2 + BC^2. \]
  We are given that $AP^2 > BP^2 + PC^2$, so $AB^2 + AC^2 > BC^2$,
  and we get $\angle BAC$ is acute.
\end{remark*}