For any triangle ABC on the plane, we draw a circle O such that AB is the diameter. We place inside circle the point P. How show with at least one of the triangles $PAB, PBC, PAC$ has a smaller perimeter than ABC? I have no idea how to do this, can this be proved with simple geometry?
hint: consider a triangle $EAC$, see picture below:
we want to prove $EB+EC < AB+AC \iff EB+OC < OE+AC \iff EB-AC < OE-OC$
Flip $C$ to $C' $ $ \implies AC=BC' ,OC=OC' ,EB-AC < OE-OC \iff EB-BC'< OE-OC'=EC'$
now prove $PB+PC <EB+EC$, you can go on now. it is easy to let $P$ on the circle first. then inside circle.