IMO 1977 LL USA54

Origin: USA

Problem

If 0 \leqa \leqb \leqc \leqd, prove that abbccdda \geqbacbdcad.

Solution

We shall use the following lemma. Lemma. If a real function f is convex on the interval I and x, y, z \inI, x \leqy \leqz, then (y −z)f(x) + (z −x)f(y) + (x −y)f(z) \leq0. Proof. The inequality is obvious for x = y = z. If x < z, then there exist p, r such that p + r = 1 and y = px + rz. Then by Jensen’s inequality f(px + rz) \leqpf(x) + rf(z), which is equivalent to the statement of the lemma. By applying the lemma to the convex function −ln x we obtain xyyzzx \geq yxzyxz for any 0 < x \leqy \leqz. Multiplying the inequalities abbcca \geqbacbac and accdda \geqcadcad we get the desired inequality. Remark. Similarly, for 0 < a1 \leqa2 \leq\cdot \cdot \cdot \leqan it holds that aa2 1 aa3 2 \cdot \cdot \cdot aa1 n \geqaa1 2 aa2 3 \cdot \cdot \cdot aan 1 .