of ( i ) . Let x ( t ) be an eventually positive solution of inequality (3) such that x ( t ) > 0 , x ( t − τ ) > 0 , x ([ t − 1]) > 0 for t > T ≥ t$_{0}$ +max { τ, 2 } , where T is sufficiently large. Set y ( t ) = ∏ T <t$_{j}$ ≤ t (1 − b$_{j}$ ) x ( t ) . Since 1 − b$_{j}$ > 0 , it is clear that y ( t ) > 0 , y ( t − τ ) > 0 , and y ([ t − 1]) > 0 for t > T. Now, we will show that y ( t ) is a solution of inequality (6). From (8), (9), and (3) we obtain that So, y ( t ) is an eventually positive solution of inequality (6). On the other hand, from (2), we have and 5 y ( t + i ) = ∏ T <t$_{j}$ ≤ t$_{i}$ (1 − b$_{j}$ ) x ( t + i ) = ∏ T <t$_{j}$ ≤ t$_{i}$ (1 − b$_{j}$ ) x ( t$_{i}$ ) = y ( t$_{i}$ ) . y ( t − i ) = ∏ T <t$_{j}$ ≤ t$_{i}$$_{−}$$_{1}$ (1 − b$_{j}$ ) x ( t − i ) = ∏ T <t$_{j}$ ≤ t$_{i}$ (1 − b$_{j}$ ) x ( t$_{i}$ ) = y ( t$_{i}$ ) = ∏ T <t$_{j}$ ≤ t (1 − b$_{j}$ ) x $^{′}$( t ) + a ( t ) ∏ T <t$_{j}$ ≤ t (1 − b$_{j}$ ) x ( t ) + ∏ t − τ<t$_{j}$ ≤ t (1 − b$_{j}$ ) b ( t ) ∏ T <t$_{j}$ ≤ t − τ (1 − b$_{j}$ ) x ( t − τ ) + ∏ [ t − 1] <t$_{j}$ ≤ t (1 − b$_{j}$ ) c ( t ) ∏ T <t$_{j}$ ≤ [ t − 1] (1 − b$_{j}$ ) x ([ t − 1]) ≤ ∏ T <t$_{j}$ ≤ t (1 − b$_{j}$ ) [ x ′ ( t ) + a ( t ) x ( t ) + b ( t ) x ( t − τ ) + c ( t ) x ([ t − 1])] ≤ 0 . y $^{′}$( t ) + a ( t ) y ( t ) + B ( t ) y ( t − τ ) + C ( t ) y ([ t − 1])