IMO 2012 Shortlist A3

Category: Algebra

Problem

Let a2,...,an be n − 1 positive real numbers, where n ≥ 3, such that a2a3 ···an = 1. Prove that (1 + a2)2 (1 + a3)3 ···(1 + an)n

nn . Solution. The substitution a2 = x2 x1 , a3 = x3 x2 , ..., an = x1 xn−1 transforms the original problem into the inequality (x1 + x2)2 (x2 + x3)3 ···(xn−1 + x1)n nn x2 1x3 2 ···xn n−1 (∗) for all x1,...,xn−1 > 0. To prove this, we use the AM-GM inequality for each factor of the left-hand side as follows: (x1 + x2)2 ≥ 22 x1x2 (x2 + x3)3 = 2 x2

x3 3 ≥ 33 x2 2 x3 (x3 + x4)4 = 3 x3
x4 4 ≥ 44 x3 3 x4 . . . . . . . . . (xn−1 + x1)n = (n − 1) xn−1 n−1
x1 n ≥ nn xn−1 n−1 n−1 x1. Multiplying these inequalities together gives (*), with inequality sign ≥ instead of >. However for the equality to occur it is necessary that x1 = x2, x2 = 2x3, ..., xn−1 = (n−1)x1, implying x1 = (n − 1)!x1. This is impossible since x1 > 0 and n ≥ 3. Therefore the inequality is strict. Comment. One can avoid the substitution ai = xi/xi−1. Apply the weighted AM-GM inequality to each factor (1 + ak)k, with the same weights like above, to obtain (1 + ak)k = (k − 1) k − 1
ak k ≥ kk (k − 1)k−1 ak. Multiplying all these inequalities together gives (1 + a2)2 (1 + a3)3 ···(1 + an)n ≥ nn a2a3 ···an = nn . The same argument as in the proof above shows that the equality cannot be attained.13