Elementary Inequalities - Tutorial Text No. 1 by D. S. Mitrinovic

By D. S. Mitrinovic

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N i=l The case of equality is easily considered. (ii) A very simple inductive proof for the equal weight case has been given by > 2 and that the result is Aczel, [Aczel1961a,b; Pecaric 1990c]. Assume that n known for all k, 2 1 f ( n n~ai ) < k < n. (n~ai) +n-1~f(ai)' 1 1 n ) 1 n- ) by the case n == 2, and the induction hypothesis. 1 Theorem 6(b), Remark (x). 1 (1) holds strictly if X # y' A # 0' 1 . With a slight change of notation we can rewrite (J) in this case as: f( 1- sx + sy) < (1- s)f(x) + sf(y), 0 < s < 1, with equality only if x == y.

N . -L((i-1)t(~- )+(n-i+1)t(~))>Lt( ~ ). n. n n . n+1 ~=1 ~=1 So n-1 . L ((i- 1)t(~-n 1 1 . 1 ) + (n- i + 1)t( ~ )) + -t(1) > n ~=1 n n+1 . L. t( n+1 ~ ) - t(1). ~=1 This on simplification gives the first inequality. A similar argument, starting with to the same inequality when REMARK (xii) 1 i f ( i + ) + (1 i )f ( i ) n+ 1 n+1 n+1 n+1 f is concave. leads D This result is a special case of a slightly more general result; [Kuang; Qi 2000a]. 122-123]. THEOREM 9 [HADAMARD-HERMITE INEQUALITY] Iff : [a, b] ~ 1R is convex and if a < c < d < b, tben c+d)< 1 1df

1 ) 1 fax (x-t)n f(n+l) (t) dt is the (n+l)-st Taylor remainder centred at a; also for some c, a

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