SOLUTIONS

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143. [2003, 201; 2004, 129] Proposed by José Luis Diaz-Barrero, Universidad Politécnica de Cataluña, Barcelona, Spain.

Let α, β, and γ be the angles of acute triangle ABC. Prove that

\frac{cot α cot β}{\sqrt{1 − cot α cot β}} + \frac{cot β cot γ}{\sqrt{1 − cot β cot γ}} + \frac{cot γ cot α}{\sqrt{1 − cot γ cot α}} ≥ \sqrt{\frac{3}{2}}.

Solution by Ovidiu Furdui, Western Michigan University, Kalamazoo, Michigan. First we notice that the expressions under the square roots are positive since

1 − cot α cot β = \frac{− cos(α + β)}{sin α sin β} ≥ 0,

since sin α, sin β > 0 and cos(α + β) ≤ 0. (i.e. α + β ≥ 90^{∘}; 180^{∘} − γ ≥ 90^{∘} implies γ ≤ 90^{∘}). I’ll make use of the following equality which holds in any triangle:

cot α cot β + cot β cot γ + cot γ cot α = 1,

where α, β, and γ are the angles of a given triangle ABC. Denote

a = cot α cot β, b = cot β cot γ, c = cot γ cot α.

Let

f: (0, 1) → ℝ, where f(x) = \frac{x}{\sqrt{1 − x}}.

Then

f′(x) = \frac{2 − x}{2(1 − x)^{\frac{3}{2}}} and f′′(x) = \frac{4 − x}{4(1 − x)^{\frac{5}{2}}} > 0.

Therefore, f is a convex function on (0, 1). By applying Jensen’s Inequality to f we get that

f(\frac{a + b + c}{3}) ≤ \frac{f(a) + f(b) + f(c)}{3}.

Therefore,

\frac{a}{\sqrt{1 − a}} + \frac{b}{\sqrt{1 − b}} + \frac{c}{\sqrt{1 − c}} ≥ \frac{3 · \frac{a+b+c}{3}}{\sqrt{1 − \frac{a+b+c}{3}}} = \frac{1}{\sqrt{1 − \frac{1}{3}}} = \sqrt{\frac{3}{2}};

since a + b + c = 1.

Also solved by Joe Howard, Portales, New Mexico; Mihai Cipu, Romanian Academy, Bucharest, Romania; and the proposer.