Let 0 < mI ≤ A ≤ m I ≤ M I ≤ B ≤ MI and p ≥ 1. Then for every positive unital linear map Φ, Φ 2p (A t B) ≤ (K(h,2) 4 1 p −1 (1+Q(t)(log M m) 2)) 2p Φ 2p (A t B) and Φ 2p (A t B) ≤ (K(h,2) 4 1 p −1 (1+Q(t)(log M m) 2)) 2p (Φ(A) t Φ(B)) 2p , where t ∈ [0, 1], h = M m , K(h, 2) = (h+1) 2 4h , Q(t) = t 2 2 (1−t t) 2t and Q(0) = Q(1) = 0. Moreover, we give an improvement for the operator version of Wielandt inequality.