Problem 11984
(American Mathematical Monthly, Vol.124, May 2017) Proposed by D. Sitaru (Romania).
Leta, b, and c be the lengths of the sides of a triangle with inradius r. Prove a6+ b6+ c6≥5184r6.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let u = b + c − a, v = a + c − b, w = a + b − c. Then u, v, w > 0 and pr =pp(p − a)(p − b)(p − c) =⇒ 2r =
r uvw
u + v + w
where p = (a + b + c)/2. Moreover 5184 = 26·34. Hence the inequality is equivalent to
v + w 2
6
+ u + w 2
6
+ u + v 2
6
≥34· u3v3w3 (u + v + w)3 or
v+w 2
6
+ u+w2 6
+ u+v2 6
3 · u + v + w
3
3
≥u3v3w3 which holds because, by the AM-GM inequality,
v+w 2
6
+ u+w2 6
+ u+v2 6
3 ≥ (vw)3+ (uw)3+ (uv)3
3 ≥ (vw)3·(uw)3·(uv)31/3
= u2v2w2, and
u + v + w 3
3
≥uvw.
