Problem 11686
(American Mathematical Monthly, Vol.120, January 2013) Proposed by Michel Bataille (France).
Let x, y, z be positive real numbers such that x + y + z = π/2. Prove that cot x + cot y + cot z
tan x + tan y + tan z ≥ 4(sin2x + sin2y + sin2z).
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let x = 2α, y = 2β and z = 2γ, where α, β, γ are the angles of a triangle. with sidelengths a, b, c.
Then triangle inequality implies that u = s − a, v = s − b, and w = s − c are non negative numbers where s = (a + b + c)/2. It is well known that
tan x =r vw
su, tan y =r wu
sv, tan z =r uv sw. Therefore
cot x + cot y + cot z = cot x cot y cot z = r s3
uvw, cot x cot y + cot y cot z + cot z cot x = s 1
u+1 v + 1
w
sin2x + sin2y + sin2z = tan2x
1 + tan2x+ tan2y
1 + tan2y + tan2z
1 + tan2z = vw
su + vw + wu
sv + uz + uw sw + uv. Hence, the required inequality becomes
(u + v + w)2
uv + vw + wu ≥ 4vw
(u + w)(u + v)+ 4wu
(v + u)(v + w) + 4uv (w + v)(w + u) or the following symmetric homogeneous inequality in the variables u, v, w ≥ 0,
(u − v)2W + (v − w)2U + (w − u)2V ≥ 0 where
U = v2(w+u)+w2(v+u)−u2(v+w), V = w2(u+v)+u2(w+v)−v2(w+u), W = u2(v+w)+v2(u+w)−w2(u+v).
Without loss of generality we assume that u ≥ v ≥ w ≥ 0, then
V = (u − v)(uv + vw + wu) + w2(u + v) ≥ 0, and W = (u − w)(uv + vw + wu) + v2(u + w) ≥ 0.
Therefore
(u − v)2W + (v − w)2U + (w − u)2V ≥ (v − w)2(U + V ) = 2(v − w)2w2(u + v) ≥ 0.
