22nd APMC 1999 Problem 2 Find the best possible k, k' such that k < v/(v + w) + w/(w + x) + x/(x + y) + y/(y + z) + z/(z + v) < k' for all positive reals v, w, x, y, z.
1, 4 Solution Put f(v,w,x,y,z) = v/(v+w) + w/(w+x) + x/(x+y) + y/(y+z) + z/(z+v). We have v/(v+w) + w/(v+w) = 1, etc, so 5 - f(v,w,x,y,z) = w/(v+w) + x/(w+x) + y/(x+y) + z/(y+z) + v/(z+v) = w/(w+v) + v/(v+z) + z/(z+y) + y/(y+x) + x/(x+w) = f(w,v,z,y,x). So k + k' = 5. Note that f(ε 4 ,ε 3 ,ε 2 ,ε,1) = 4ε/(1+ε + 1/(1+ε 4 ) → 1 as ε → 0. So k ≤ 1. So it remains to prove that f(v,w,x,y,z) > 1 for all v,w,x,y,z. It is sufficient to prove it for v,w,x,y,z such that v+w+x+y+z = 1, since f(kv,kw,kx,ky,kz) = f(v,w,x,y,z). So assume v+w+x+y+z=1. 1/x is convex for positive, so by Jensen's inequality v/x 1 + w/x 2 + x/x 3 + y/x 4 + z/x 5 ≥ 1/(vx 1 + wx 2 + xx 3 + yx 4 + zx 5 ). Putting x1 = v+w, x2 = w+x etc we get f(v,w,x,y,z) ≥ 1/∑v(v+w). But 1 = (v+w+x+y+z) 2 > v(v+w) + w(w+x) + x(x+y) + y(y+z) + z(z+w), so f(v,w,x,y,z) > 1, as required. Hence k = 1. Hence k' = 4. 22nd APMC 1999 © John Scholes jscholes@kalva.demon.co.uk 28 Nov 2003 Last corrected/updated 28 Nov 03