By ed.B.Pachpatte

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**Example text**

4), for j = m are obviously satisfied if 1 ~< k ~< m and r ~< k ~< n. 4) is invalid, that is, Y~I=I Pl (Xl --Xm) ~ O. 7), we have y~,n=k ~+l Pl (Xl -- Xm ) ~ O, w e get Y~,n__1 Pz (x~ - Xm) >~ O, that is, s ~> Xm, what is evidently a contradiction. 2) holds. If Xl ~< . . 2) is valid. 2) holds. Let f (x) = x 2, Xl = 0, i = 1 . . . k - 1, and 1, i = k . . . n. 2) becomes (ffk / pn)2/> ffk / Pn. Hence, ~ ~< 0 or Pk-1 ~<0, k = 2 . . . n. X ! -'- Now l e t k < m and suppose that Pk~<0. L e t x t = 0 , l ~ < i < ~ k - l , x ~ = l , k ~< i ~< m - 1, and Xl - 1 + e, m <~ i <<,n.

I <~ n, such that e. 11) 9 PROOF. Consider the functional 1s J ( f ) = ~n ,=l P' f (x') - f (1s -~n i=lP'X' , f e F(I). Then J satisfies conditions (J1) and (J2) (by Jensen's inequality). 11). The proof is finished. Fq In [88] the following improvement of Fuch's generalization of the majorization theorem (see [122]) is given. 4. Let al >/ ... >/as, bl >/... >1 bs and ql . . . s-1, t=l ~q,a,=~q,b,. t=l t=l If g is convex on I and f is g-convex dominated on I, then the following inequality holds: ~q,(f(b,)t--I f(a,)) <~~ i=1 (g (b,) - g (ai)).

For any function gl :E --+ ~ such that kgl ~ L and kq~(gl) E L, we have dp A(kgl) ) <~ A(kdp(gl)) . 11) If in addition, I = [m, M] where - o o < m < M < oo, then A[k~hrglah~,z~ J! 3, respectively. PROOF. In case gl E L and ~(gl) E L, and k is such that kh ~ L for all h E L, the functional F :L -~ R defined by F(h) = A(kh) A(k) ' h ~ L, 40 Chapter 1. Inequalities Involving Convex Functions is an isotonic linear functional satisfying F (1) = 1. 4). 4). 11) since similar modifications handle the other proofs.

### Mathematical Inequalities by ed.B.Pachpatte

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