ON THE ARITHMETIC AND GEOMETRIC MEANS AND ON HOLDER'S INEQUALITY H KOBER The theorem on the arithmetic and geometric means can be sharpened; in consequence of this, Holder's inequality can be sharpened also, and finally, Minkowski's inequality It is well known that the difference between the two means, (1.1) D„ = (xi + x2 + • • • + x„)/n is the square of an irrational w>2 For instance1 n V1 Da = 2-, I I 1/3 _L \ix\ function 1/3 J + x2 — (xix2 ■ ■ ■ xn)lln, for n = 2, a sum of squares 1/3,1/2 + Xi ) 1/3 (Xi 1/3 — xk )/6 for 1/2,2 } lS>0, unless = A, i.e a\ = a2= ■ ■ ■ =an+i; which case, however, was excluded Thus g„+i(a*)>0 unless = Now we consider gn+i(a) —gn+iia*) n+l dn+i = (n- We have / n+l l)Z^-2(a1)1'2( =2 -2 =dn+i \ £ \ Z («i)1/2 - n(A)1'2) i-2 / ('2 - n(AyA \ Since (a2a3 gn(a2, >'=2 - n\l / a3, • • • , an+i) ^ by assumption, ■ ■ ■ an+i)lln=A, Loc cit., Theorem 9, Inequality (2.5.2) License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use a)"\ »-2 ^ A and 1958] ARITHMETIC AND GEOMETRIC MEANS n+1 dn+i * £ n+1 455 n+1 - 2(AY'2 E (at)1'* + nA = }2 ((« n D„—A„/(nqn) ^0, 22 ((Xi)l/2 ~ (Xk)il2)2 ^ qn If q2 = qz= ■ ■ -=qn>qi the first term on the right in (4.2) reduces to (qn —qi)(nqn)~lXi; therefore Dn— An/(nqn) =0 if, and only if, Xi = xqi1x2> ■ • ■x„" Combining this with the above condition resulting from the lemma we deduce that Xi = 0, x2 = Xz= ■ ■ •=x„>0, which completes the proof On Holder's inequality We employ a well-known argument.4 Set aik/sk = bik; then bik-\-b2k-\- ■ ■ • -\-bmk= l for each k Now m m / n \ 22 bnbu ' bin= 22 ( ?i^ô'i+ 1ibi2+ • • • —qnbin+ II bikJ t-i t-i \ *=i / We can apphr Theorem to each of the expressions Loc cit §2.7, proof (ii) of Holder's inequality License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use in brackets Thus 456 H KOBER Oi m -^— E (5.1)B"UuiWs" E ((&«)1/J- (ft*)1'*)1 gl-EIIJ^ min^ 1,9„E t—1 fc—1 which [June is equivalent V E i=l to the result ((M1/2" (M1/2)4 , lsj'S, unless aik/pk = an/pi for any i, k, I, i.e unless the at], an, ■ ■ ■ are propor- tional The proof is based bik = a\k, then i (6.2) S=S on a well-known m m ^ If o-;= E"-i m aô> i~l < E a'l'i + E 0 is fixed; for wS:2, there is never a unique limit when Xo= 0, except for the trivial case n = 2, qi = q2=l/2 We set Un(x) = Z((x;)1/2 - (xk)1'2)2 (1 ^ i < k S n) and take 22i ixi —Xo)2—>0.We observe that not all the Xi are equal I Let n>2, (1.1) qi 2(« - l)-1^^! -