IMO 1979 SL 10

Origin: FIN

Problem

Show that for any vectors a, b in Euclidean space, |a \times b|3 \leq3 \sqrt 8 |a|2|b|2|a −b|2. Remark. Here \times denotes the vector product.

Solution

In the cases a = −\to0 , b = −\to0 , and a \parallelb the inequality is trivial. Otherwise, let us consider a triangle ABC such that −−\to CB = a and −\to CA = b. From this point on we shall refer to \alpha, \beta, \gamma as angles of ABC. Since |a \times b| = |a||b| sin \gamma, our inequality reduces to |a||b| sin3 \gamma \leq3 \sqrt 3|c|2/8, which is further reduced to sin \alpha sin \beta sin \gamma \leq3 \sqrt using the sine law. The last inequality follows immediately from Jensen’s inequality applied to the function f(x) = ln sin x, which is concave for 0 < x < \pi because f ′(x) = cot x is strictly decreasing.