a, b, c are fixed positive reals. Find all positive real solutions x, y, z to: x + y + z = a + b + c and 4xyz - (a2x + b2y + c2z) = abc.
Solution
Answer: the unique solution is x = (b+c)/2, y = (c+a)/2, z = (a+b)/2.
Dividing by xyz, the second equation becomes 4 = a2/yz + b2/zx + c2/xy + abc/xyz. Put X = a/(yz)1/2, Y = b/(zx)1/2, Z = c/(xy)1/2, then this becomes 4 = X2 + Y2 + Z2 + XYZ (*). All of a, b, c, x, y, z are positive, so X, Y, Z are positive. Hence X2 < 4, so X < 2. Similarly, Y and Z < 2. If we regard (*) as a quadratic in Z, then we can solve it to get Z = ( (X2Y2 + 16 - 4X2 - 4Y2)1/2 - XY)/2 (the other solution is negative and hence ruled out).
The key is to subsitute X = 2 sin u, Y = 2 sin v, then Z = 2(cos u cos v - sin u sin v) = 2 cos(u + v). So we have a = 2 (yz)1/2 sin u, b = 2 (zx)1/2 sin v, c = 2 (xy)1/2(cos u cos v - sin u sin v). So far we have not used the first equation. If we require that it holds then we have x + y + z = 2 (yz)1/2 sin u + 2 (zx)1/2 sin v + 2 (xy)1/2 (cos u cos v - sin u sin v) (**). Writing x = ( x1/2 sin v)2 + (x1/2 cos v)2, y = (y1/2 sin u)2 + (y1/2 cos u)2, z = (z1/2)2, we can write (**) as (x1/2 cos v - y1/2 cos u)2 + (x1/2 sin v + y1/2 sin u - z1/2)2 = 0. Squares are non-negative, so each term must be zero. Hence z1/2 = x1/2 sin v + y1/2 sin u = 1/z1/2 (b/2 + a/2). Hence z = (a + b)/2.
But the problem is symmetrical. So we must also have x = (b + c)/2, y = (c + a)/2. It is not hard to check that this is indeed a solution. We get 8xyz - 2(a2x + b2y + c2z) = (b + c)(c + a)(a + b) - (a2(b+c) + b2(c+a) + c2(a+b) ) = 2abc (after some cancellation).
© John Scholes
jscholes@kalva.demon.co.uk
4 Sep 2002