Page 38 - RMGO 6
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            38                                                        Dorin MARGHIDANU

                On the other hand, using n times the right inequality from Lemma 1, we obtain
                         n                            n                   n
                     1  X   x k |x k | − x k+1 |x k+1 |  1  X            X
                       ·                        ≤   ·    (|x k | + |x k+1 |) =  |x k | .
                     2           x k − x k+1      2
                        k=1                          k=1                 k=1

            Remark 2. For n = 2 and n = 3 in (5), we regain the inequalities (1) and (4),
            respectively.


            References


             [1] D. M˘arghidanu, A refinement of modulus inequality, proposed problem in Short
                Mathematical Idea, https://www.facebook.com/photo/?fbid=6474936872565366&
                set=gm.2819058648230558&idorvanity=370199313116516

             [2] D. M˘arghidanu, A refinement of modulus inequality (the general case), proposed
                problem in Mathematical Inequalities, https://www.facebook.com/photo?fbid=
                6499590363433350&set=gm.3461865910768112&idorvanity=1486244404996949
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