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									An Unsettled Inequality: 10337

         Horst Alzer; M. J. Pelling

         The American Mathematical Monthly, Vol. 106, No. 10. (Dec., 1999), p. 970.

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                                                                                                      Sat Dec 22 05:46:31 2007
REVIVALS 

                                     A n Unsettled Inequality
10337 [1993, 798; 1995, 6591. Proposed by Horst Alzel; Waldbrol, Germany. Let n 2 1
be an integer. Let x l , . . . , x, be real numbers with xi E (0, 1/21. Consider the statement



( a )Prove F, for n 5 3.
(b) Show that F, is false for n 2 6.
(c)* What about Fq and F5?
Solution of part (c)* by M. J. Pelling, London, England. We show that F4 is true, with
equality if and only if xl = x2 = x3 = x4.
   Write w , x , y, z for x l , x2, x3, xq, and write 6 ,i ,j , i for 1 - w , 1 - x , 1 - y, 1 - z ,
respectively. Then F4 may be written in the equivalent form
                          w4                           +
                               + x4 + y4 + z4 2 w4 + i4 j4 + i4
                                wxyz                wxjz
Without loss of generality, suppose that w 2 x 2 y 2 z. Subtracting 4 from both sides of
( 1 ) and rearranging terms leads to
(w2 - x2)2
              + ( y 2 - z2)2   +   ~ ( W -
                                         X   yz)2   , -i212
                                                    -
                                                     (w2              ( j 2 - i212 2(Wi - ji12
                                                                                 +
   WXYZ             WXYZ              WXYZ              dPji     +        wiQi        WXjZ     .
                                                                                             (2)
By repeated use of the elementary inequality
                                 1        1
                                         +
                         p + - 2 q - wheneverp 2 q 2 1,
                                 P        4
                                                                                              (3)
we show that each term on the left of (2) is greater than or equal to the corresponding term
on the right.
   Since w + x i 1 , w e h a v e w - x 2 w 2 - x 2 0 r w w > x i . Withp = w l x a n d q = i / G ,
wehave p p q 2 1, so
                                          +
                                      ( w x ) ~ (W x ) 2
                                                2
                                                        +                                     (4
                                         WX           wi
by (3). Since yz 5    72 and ( w - x                    ,
                                        ) = (W - z ) ~( 4 ) implies
                                          ~



The same reasoning proves

                                                    >
                                                    -
                                     wxyz         wiyz ,
Now let p = w x / ( y z ) and q = jil(w.2). Again p 2 q 2 1, so (3) implies
                                      2(wx - y z ) 2       2(GX - y z ) 2
                                                         2                                    (7)
                                            wxyz               wxyi .
Adding (5), (6),and ( 7 ) yields (2).
     Since equality holds in ( 3 )only when p = q , we have equality in F4 only if w / x = i / G ,
y / z = i / j , and w x / ( y z ) = j i / ( G i ) , which forces w = x = y = z.
Editorial comment. Pelling also contributed a lengthy proof of Fg and showed that equality
holds in F5 only when X I = x2 = x3 = x4 = xg.

970                                 PROBLEMS AND SOLUTIONS                           [Monthly 106

								
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