IMO 1990 SL 24

Origin: THA

Problem

Let a, b, c, d be nonnegative real numbers such that ab + bc + cd + da = 1. Show that a3 b + c + d + b3 a + c + d + c3 a + b + d + d3 a + b + c \geq1 3.

Solution

Let us denote A = b + c + d, B = a + c + d, C = a + b + d, D = a + b + c. Since ab + bc + cd + da = 1 the numbers A, B, C, D are all positive. By trivially applying the AM-GM inequality we have: a2 + b2 + c2 + d2 \geqab + bc + cd + da = 1 .

We will prove the inequality assuming only that A, B, C, D are positive and a2 + b2 + c2 + d2 \geq1. In this case we may assume without loss of generality that a \geqb \geqc \geqd \geq0. Hence a3 \geqb3 \geqc3 \geqd3 \geq0 and A \geq1 B \geq1 C \geq D > 0. Using the Chebyshev and Cauchy inequalities we obtain: a3 A + b3 B + c3 C + d3 D \geq1 4(a3 + b3 + c3 + d3)  1 A + 1 B + 1 C + 1 D  \geq1 16(a2 + b2 + c2 + d2)(a + b + c + d)  1 A + 1 B + 1 C + 1 D  = 1 48(a2 + b2 + c2 + d2)(A + B + C + D)  1 A + 1 B + 1 C + 1 D  \geq1 3 . This completes the proof.