Before continuing, let's note that as #x, y, z > 0# we have #a, b, c >0# and so we may take square roots of any product or quotient thereof, and do so without considering negative solutions.
#{(xy = a),(yz = b),(xz=c):}#
From the first equation, we have
#y = a/x#
Substituting this into the second equation:
#b = yz = a/xz#
#=> z = b/ax#
Substituting this into the third equation:
#c = xz = b/ax^2#
#=> x^2 = (ac)/b#
#=> x = sqrt((ac)/b)#
Substituting our solution for #x# into our intermediate result from working on the first equation:
#y =a/x = a/sqrt((ac)/b) = sqrt((ab)/c)#
Substituting our solution for #x# into our intermediate result from working on the second equation:
#z = b/ax = b/asqrt((ac)/b) = sqrt((bc)/a)#
#:.{(x = sqrt((ac)/b)), (y = sqrt((ab)/c)), (z = sqrt((bc)/a)):}#
