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2pr/R <=DE + EF + DF <= p, touchpoints with incircle

Source: IV All-Ukrainian Tournament of Young Mathematicians, Qualifying p10

May 19, 2021
geometryincirclegeometric inequalityUkrainian TYM

Problem Statement

Given a triangle ABCABC and points D,E,FD, E, F, which are points of contact of the inscribed circle to the sides of the triangle.
i) Prove that 2prR≤DE+EF+DF≤p\frac{2pr}{R} \le DE + EF + DF \le p (pp is the semiperimeter, rr and RR are respectively the radius of the inscribed and circumscribed circle of △ABC\vartriangle ABC).
ii). Find out when equality is achieved.
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