This can be shown in three steps.

Show that every closed borken line of perimeter 1 can be enclosed by a circle of radius 1/4. (Chose two points on the line that divide the perimeter into two halves. Then consider a circle with the two points defining a diameter.)

Show that if a triangle is entirely inside a circle C of radius S then the circumradius R < S. (The arc of the circumcircle that may be outside C is less than 180°, because the triangle is acute. Therefore, there is a circumdiameter which is entirely inside C.)

By 1, a triangle of perimeter p can be encircled by a circle of radius p/4. By 2, its circumradius R satisfies R < p/4. If 2s = p, you get 2R < s.