IMO shortlist 1985/4 solution

26th IMO 1985 shortlist

Problem 4

Show that x₁²/(x₁² + x₂x₃) + x₂²/(x₂² + x₃x₄) + ... + x_n-1²/(x_n-1² + x_nx₁) + x_n²/(x_n² + x₁x₂) ≤ n-1 for all positive reals x_i.

Solution

Solution by Demetres Christofides

The inequality would not have any obvious meaning if n = 1. For n = 2 it presumably becomes: a²/(a² + ba) + b²/(b² + ab) ≤ 1. But the lhs is a/(a+b) + b/(a+b) = 1, so the result is true. Assume then that n > 2.

Put a_i = x_i+1x_i+2/x_i² (using cyclic subscripts). We have to show that ∑ 1/(1 + a_i) ≤ n-1, or equivalently that ∑ a_i/(1 + a_i) ≥ 1. Also we have a₁a₂ ... a_n = 1. We may assume that a₁ ≥ a₂ ≥ ... ≥ a_n.

f(x) = x/(1+x) is an increasing function of x and f(1) = 1/2. So if there are two or more a_i ≥ 1, then ∑ a_i ≥ 1. Not all the a_i can be less than 1, because their product is 1, so we must have exactly one a_i ≥ 1 and that must be a₁. Since the others are < 1 and the product is 1, we must have a₁ > 1.

Now by the AM/GM inequality we have a₂/(1 + a₂) + ... + a_n/(1 + a_n) ≥ (n-1) ( ∏₂ⁿ a_i/(1 + a_i) )^1/n-1. But a₂a₃ ... a_n = 1/a₁ > 1/(1 + a₁) and (1 + a_i) < (1 + a₁) for i = 2, 3, ... , n-1, so (n-1) ( ∏₂ⁿ a_i/(1 + a_i) )^1/n-1 > (n-1) 1/(1 + a_n)^1/n-1 1/(1 + a₁). Now a_n < 1, so (n-1)/(1 + a_n)^n-1 > 2/2^1/n-1 > 1. Hence a₂/(1 + a₂) + ... + a_n/(1 + a_n) > 1/(1 + a₁). Hence a₁/(1 + a₁) + ... + a_n/(1 + a_n) > 1.

26th IMO shortlist 1985