Prove that the system $$\sum_{i=1}^{m} x_i a_i = 0, x_i > 0, i = 1, . . . , m,$$
has no solution if and only if the system $$<a_i , y> ≤ 0, i = 1, . . . , m,$$ not all zero has a solution. Hint: Use the finite-dimensional version of the Dubovitskii–Milyutin theorem.
Proof
Let $K_i=\{x:x_i >0\}, i=1,...,m$, $K_{m+1}=Ker(A)$ where $A=[a_1,...,a_m]$.