Let $ABCD$ be a parallelogram. Construct the equilateral triangle $DCE$ on the side $DC$ and outside of parallelogram. Let $P$ be an arbitrary point in plane of $ABCD$. Show that
\[PA+PB+AD \geq PE.\]
rotation with angle $\angle 60$ with center $E$ (counterclock),transforms point $P$ to $P'$.$P'E=PE,ED=EC,\angle DEP'=\angle PEC$ so $P'D=PC$.rotation with angle $\angle 60$ with center $C$ (counterclock) transforms point $B$ to $B'$.$P'D=PC,AD=CB',\angle P'DA=\angle PCB'$ so $PB'=AP'$.
$PE=PP'\le P'A+PA=PB'+PA\le PB+BB'+PA=PA+PB+AD$