Problem 12168
(American Mathematical Monthly, Vol.127, March 2020) Proposed by M. Lukarevski (North Macedonia).
Leta, b, and c be the side lengths of a triangle ABC with circumradius R and inradius r. Prove 2
R ≤ sec(A/2)
a +sec(B/2)
b +sec(C/2)
c ≤ 1
r.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let s be the semiperimeter of the triangle and let F be its area. Then we have that
r = F
s, R = abc
4F, sec(A/2) =
s bc
s(s − a), sec(B/2) =
r ca
s(s − b), sec(C/2) =
s ab
s(s − c), and the double inequalities can be written as
8F2
abc ≤pbc(s − b)(s − c)
a +pca(s − c)(s − a)
b +pab(s − a)(s − b)
c ≤ s.
We first show the inequality on the right: by AM-GM inequality pbc(s − b)(s − c)
a ≤
b+c
2 ·2s−b−c2
a =b + c 4 and therefore
pbc(s − b)(s − c)
a +pca(s − c)(s − a)
b +pab(s − a)(s − b)
c ≤ b + c
4 +c + a
4 +a + b 4 = s.
As regards the inequality on the left, let x = 12(b + c − a) > 0, y = 12(c + a − b) > 0, and z = 12(a + b − c) > 0 then
a = x + y, b = y + z, c = z + x, s = x + y + z, F2= xyz(x + y + z) and the inequality becomes
8xyz(x + y + z) ≤ RHS where
RHS :=py(y + x)3pz(z + x)3+px(x + y)3pz(z + y)3+px(x + z)3py(y + z)3. We have that
(2x + y)(x − y)2⇔py(y + x)3≥ y(y + 3x)
√2 whence
RHS ≥ yz(y + 3x)(z + 3x)
2 +xz(x + 3y)(z + 3y)
2 +xy(x + 3z)(y + 3z) 2
= x2(y − z)2+ y2(x − z)2+ z2(x − y)2
4 + 8xyz(x + y + z)
≥ 8xyz(x + y + z)
and we are done.
