 Solution

The point $(x_1,x_2,x_3)=(4,2,2)$ is nonnegative and makes all three inequalities equalities, with objective value $20$. To certify global optimality, combine the first upper bound with coefficient $1/2$, the second with coefficient one, and the lower bound with coefficient $-3/2$:This is a weak duality certificate for the linear program. The feasible point attains the bound, so the optimal solution is $(4,2,2)$ and the maximum is $20$. Equality forces each of the three constraints to be active; their simultaneous equations have this unique solution.