Indeed b > 0 is a physical constraint that must be satisfied. This constrains the domain of the cost function. And it must be minimized in that domain.
For a general multivariate function, global minimization is a very difficult problem. In this case, however, the problem can be analyzed with elementary means. Let's take some value k and see if the cost function can be less than that: i.e.,
(x^2 + z^2 +16)/(5*x+z+12) < k
(x^2 + z^2 +16) < k(5*x+z+12)
(x^2 + z^2 +16) - k(5*x+z+12) < 0
x^2 - 5kx + (2.5k)^2 - (2.5k)^2 + z^2 - kz + (0.5k)^2 - (0.5k)^2 + 16 - 12k < 0
(x - 2.5k)^2 - (2.5k)^2 + (z - 0.5k)^2 - (0.5k)^2 + 16 - 12k < 0
(x - 2.5k)^2 + (z - 0.5k)^2 - 6.5k^2 - 12k + 16 < 0
It is obvious that the minimum of the left-hand side expression is -6.5k^2 - 12k + 16, so -6.5k^2 - 12k + 16 < 0. What does this inequality mean? It means that for any k satisfying it, there are lesser values of the cost function, i.e., this value is not minimal.
There are two ranges of k satisfying this inequality: ## k > 4 \frac {\sqrt{35} - 3 } {13} ## and ## k < -4 \frac {\sqrt{35} + 3 } {13} ##. The second range is invalid, because we require that the cost-function be positive, so we have only one range, and that means that ## \displaystyle k = 4 \frac {\sqrt{35} - 3 } {13} ## is the greatest value, for which no lesser value exists, i.e., it is the global minimum. And it is exactly the value you obtained for the local minimum, so the local minimum is also global.
Two wrongs made a right!
