1. The Problem For all positive real values of a,b and c such that a²+b²+c²=1, prove that a(b+c) < [1/sqrt(2)] 2. My Attempt... There are probably numerous ways to solve this, but i tried this way... let a = sin(x) b = cos(x)sin(y) c = cos(x)cos(y) a²+b²+c² = sin²(x)+cos²(x)[sin²(y)+cos²(y)] = 1 This satisfies the above so a(b+c) = sin(x)cos(x)[sin(y)+cos(y)] =(1/2)sin2(x)[sin(y)+cos(y)]... I am sure you can somehow prove that the max value of the give is less than than the value given in the question i.e. 1/sqrt(2)... I would appreciate any help... with any other methods.