On Improvements of Kantorovich Type Inequalities

In the paper, we give some new improvements of the Kantorovich type inequalities by using Popoviciu’s, Hölder’s, Bellman’s and Minkowski’s inequalities. These results in special case yield Hao’s, reverse Cauchy’s and Minkowski’s inequalities.

The first aim of this paper is to give a new improvement of the Kantorovich type inequality (3).We combine organically Popoviciu's, Hölder's, and Hao's inequalities to derive a new inequality, which is a generalization of Label (3).
Corresponding to (3), we can obtain a reverse Minkowski's inequality as follows: where p, q, ω k , u k , v k are as in (3), and is definied in (4).Another aim of this paper is to give a new reverse Minkowski's inequality.We combine organically Bellman's and Minkowski's inequalities to derive a new inequality, which is generalization of the reverse Minkowski's inequality (5).

Results
We need the following Lemmas to prove our main results.Lemma 1. (Popoviciu's inequality) ( [18], p. 58) Let p > 0, q > 0, 1  p + 1 q = 1, and a = {a 1 , . . ., a n } and b = {b 1 , . . ., b n } be two series of positive real numbers and such that a with equality if and only if a = µb, where µ is a constant.
with equality if and only if a = υb, where υ is a constant.
Proof.From (3), we have This proof is complete.
Our main results are given in the following theorems.
where is as in (4).
Proof.Let's prove this theorem by mathematical induction for m.First, we prove that (10) holds for m = 1.From ( 3) and ( 8), we obtain and From ( 11), (12) and, in view of the Popoviciu's inequality, we have . This shows (10) right for m = 1.Suppose that (10) holds when m = r − 1; we have From ( 6), ( 12) and ( 13), we obtain . This shows that ( 10) is correct if m = r − 1, then m = r is also correct.Hence, ( 10) is right for any m ∈ N + .This proof is complete.
Taking m = 1 and ω k = µ k in Theorem 1, we have the following result.
Taking for a k = 0 and b k = 0 in (14), we get the following interesting reverse Cauchy's inequality.
where is as in (4).
Proof.First, we prove that (15) holds for m = 1.From (9) and in view of Minkowski's inequality, it is easy to obtain From ( 16), (17) and the Bellman's inequality, we have . This shows that (15) holds for m = 1 Supposing that (15) holds when m = r − 1, we have From ( 17), (18) and by using the Bellman's inequality again, we obtain This proof is complete.
Taking for m = 1, p = 2 and ω k = 1, we have the following result.

Corollary 2 .
Let u k , v k , a k and b k are as in Theorem 1, then n (15) is correct if m = r − 1, then m = r is also correct.Hence, (15) is right for any m ∈ N + .

Corollary 3 .
Let u k , v k , a k , b k , m 1 , m 2 , M 1 ,and M 2 be as in Theorem 2, then n ∑ k=1 h