If x, y, z are reals such that 0 ≤ x, y, z ≤ 1, show that x/(y + z + 1) + y/(z + x + 1) + z/(x + y + 1) ≤ 1 - (1 - x)(1 - y)(1 - z).
Solution
Consider x/(y+z+1) + y/(z+x+1) + z/(x+y+1) + (1-x)(1-y)(1-z) as a function of x, with y and z fixed. Each term is convex, so the whole function is convex. Hence its maximum value occurs at its endpoints. The same is true for x and y, so we need only check the eight possible values x, y, z = 0 or 1. In fact, we easily find the expression has value 1 at all eight points. The result follows.
A function is convex if for any three points a < b < c, the point (b, f(b) ) lies on or below the chord joining the points (a, f(a) ) and (c, f(c) ). Analytically, this means that if b = ha + (1-h)c, where 0 ≤ h ≤ 1, then f(b) ≤ h f(a) + (1-h) f(c). The linear nature of this relation implies immediately that a sum of convex functions is convex and that a positive multiple of a convex function is convex. Linear functions are obviously convex. It is obvious from the graph that the function a/(b+x) is convex. To prove it analytically we must show that a/(b+hx+(1-h)y)) ≤ ha/(b+x) + (1-h)a/(b+y) or a(b2 + bx + by + xy) ≤ ha(b+y)(b+hx+(1-h)y) + (1-h)a(b+x)(b+hx+(1-h)y). After cancelling some terms, we have to show that xy <= h(1-h)x2 + (1-h)2xy + h2xy + h(1-h)y2. This is obviously true for h = 1 or 0. Otherwise we may divide by 1-h, then h to get 2xy ≤ x2 + y2, which is true.
To see that the maximum value of a convex function must occur at its endpoints just draw a chord between the endpoints. All other points of the curve must lie below the chord.
© John Scholes
jscholes@kalva.demon.co.uk
20 Aug 2002