ON THE ARITHMETIC AND GEOMETRIC MEANS AND
ON HOLDER'S INEQUALITY
H. KOBER
1. 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><fcs3
These representations
are increasingly complicated with increasing n.
But is there not a simple sum of squares that is comparable2 with Dn
in the Hardy-Littlewood-Polya
sense? In fact, £((#;)1/2 — (xk)112)2is,
as will be shown.
Throughout
we write £««
Theorem
this paper we suppose that not all the Xi be equal; and
or £"m»a when l^i<k^n.
We prove
1. Let n^2,
0<qi^q2^
-i
..
(1.2)
qi(n — 1)
q
■ ■ • ^qn, £2»-=l,
Xi^tO. Then
An
^ -==-
12 iixi)1'2 - ixky*2y
An — 2-1 QiXi
^ qn;
Xl X2 • • • Xn .
1
If qi = q2= ■ ■ ■ =qm for some m, l^m^n,
then the lower bound
qi(n — l)-1 is reached if and only if any one of the Xi with l^i^m
is positive and the other n — 1 numbers Xi vanish; except for the trivial
case n = 2, qi = q2 = l/2. If qi<q2 = qi = • • • =qn then the upper
bound qn is attained
if and only if Xi = 0 and x2 = x$= ■ ■ ■ =xn>0.
If qi = q2= ■ ■ ■ =qn = l/n
and «^3
then the upper bound
1/n is
Received by the editors June 25, 1957 and, in revised form, October 20, 1957.
1 G. H. Hardy, J. E. Littlewood and G. P61ya, Inequalities, Cambridge, 1934, see
§2.23.
2 Loc. cit., §§1.6, 3.4, where one-sided comparability
is introduced.
Here doublesided comparability
of two functions / and g is considered, i.e. two inequalities
Og/gcig
and OggSa/,
where ci and c2 are positive constants.
452
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ARITHMETIC AND GEOMETRIC MEANS
453
attained
if and only if any one of the x,- is zero and the other m — 1
numbers
xt are positive
Remarks.
and equal.
For (b), cf. Appendix.
(a) It is an open question whether q» is the
the general case.
(b) When the point
{xi, x2, ■ ■ ■ , xn} in
(w>2) tends to {x0, x0, • ■ ■ , x0}, where Xo>0,
tends to a unique limit if, and only if, qi = q2=
limit is 2m~2.
Concerning Holder's inequality we deduce
Theorem
i = l,
2. Let 0<^i^g2^
2, ■ ■ ■ , m; j=l,
= sk>0fork
• • • ^?»i
least upper bound in
M-dimensional
space
the fraction in (1.2)
• • ■ =qn=l/n.
This
Xs* = l and a,jt>0;
2, • ■ ■ , n; m, m^2;
aik+a2k+
■ ■ ■ +amk
= l, 2, • • • , m. Then (cf. 5.1)
m
anau
■ ■ ■ ain g
max
{0; Si s2 • ■ ■ sn (1 — qnL)\,
anai2
■ • • ain S sx s2 ■ ■ • sn [ 1-L
i—l
(1.3)
»=i
\
n —1 /
I,
where
L-t z {(T-(-)'!•
•=1 lsK*Sn
If m = 2
and
\\Sj/
\ Sk/
qi = q2 = 1/2,
then
J
trivially
22(anai2)xl2
= (sis2y'2(l-L/2).
The corresponding
problem for Minkowski's
inequality
is more
difficult, it plays a part in an investigation
into arc length and surface
area. One result is given in §6.
2. On the lower bound in Theorem
An- E((*<)1/,-(«fc)1/,)V(«-l)-1=*M
1. First take qi<q2 and set
Then
n
n
dn = 22 (ii — ?l)*»' + {2?l(» — I)"1} 22 (XiXk)
— XiX2 ■ ■ ■X9n.
But
^^
n(n — 1)
22 (Of~ ?0 +-29l(n
Using the Theorem
- I)"1 = (1 - nqi) + nqi = 1.
on the arithmetic
and geometric
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means
in its
454
H. KOBER
generalized
form3 we deduce
x2 = xa =
•••
= *»=
that dn>0,
ixix2)112
=
[June
unless
(xix-,)112 =
• • • =
(x2x3)112 =
• • • ,
i.e. unless xx>0 and x2 = x3 = ■ ■ ■ =xn = 0, since not all the Xi are
equal. In a similar way we deal with the cases qi = q2<q3, qi = q2
= q3<qi,
etc.
3. Now we need a lemma.
If fl;^0, n>2, not all the a, are equal and
n
(3.1)
n
gn(a) = (n-2)22«i+
n(aia2 ■ ■ ■ a.)1'" - 2 £
(a.a*)1'2,
f=i
<Aewg«(a) > 0, unless one of the a, is zero and the others are positive and
equal.
Without
loss of generality
we assume that ax^a2^
■ ■ ■ ^a„.
Trivially g2(a)=0. Suppose it be true that, for some n>2, gn(a)^0
for all (ai, a2, ■ ■ ■ , an), a^O.
We consider gn+i(a) and set a*=ai,
a*=a* = ■ ■ ■ =a*+i=A=(a2a3
■ ■ ■ a„+i)1/n. Then
gn+1(a*)
= (n-
l)ai+
(«+
l)a1/<"+l)^n/(n+1)
If Ci = 0 then gn+i(a*) =0 also. Let now ai>0.
gn+l(d
)
n/(n+l)
2naU(n+l)
= ?iai
*
n/(n+l)
-
Then
9ln/(n+l)
+ q2A
2W(al^)1'2.
o2n/(n+l)
— di
A
,
1
where qi = (n—l)/(2n),
q2 = (n + l)/(2n);
qi+q2=l.
Hence gn+i(a*)
>0, unless ai = A, i.e. a\ = a2= ■ ■ ■ =an+i; which case, however, was
excluded. Thus g„+i(a*)>0
unless ai = 0.
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
/
(<W1/2
2s«*£n+l
n+l
= gniaz, a3, • • ■ , an+i) + 22 a»
.'-2
- 2(a1)1'2( £ (ai)>'2 - n(AyA
\
Since
(a2a3
gn(a2,
>'=2
- n\l
/
a3, • • • , an+i) ^ 0
by
assumption,
■ ■ ■ an+i)lln=A,
8 Loc. cit., Theorem 9, Inequality
(2.5.2).
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a)"\
»-2
ai ^ A
and
1958]
ARITHMETIC AND GEOMETRIC MEANS
n+1
dn+i * £
n+1
455
n+1
ai - 2(AY'2 E (at)1'* + nA = }2 ((«<)1/2- (A)1'2)2.
»=2
»'=2
»'—2
This is positive, unless at = A (i = 2, 3, ■ ■ ■ , m+ 1), i.e. a2 = a3 = ■ ■ ■
= an+i- Combining the two results we complete the proof.
4. On the upper bound in (1.2). The denominator
£((*,)1/2
(4.1)
"
= nDn+
is
(Xk)112)2
{(n-2)22*i+
n(xix2 • • • *»)1/B -
2£
(x^)1'2}.
By the lemma, the term in the curled brackets is positive, unless
one of the Xi is zero and the others are equal. Under this restriction,
therefore, the fraction in (1.2) is smaller than An(nD„)~l.
(i) When qi = q2= • ■ ■ =qn = l/n clearly An = Dn, which completes the proof for the familiar case,
(ii) If not all the qi are equal, not all the differences qn —q%vanish.
Since
(4.2) Dn-
A
=-
nqn
1 n_1
nqn
1
"
nqn
i=i
22 (in - qi)Xi-\-II
,-=i
•
n
a
*i ~ II Xi
,-=i
and
"^
In-
2_,-1-=
,-_i
qi
nqn
1
l;-1-=
nqn
An/
qi t
nqn
we can use the Theorem
on the
therefore An/(nDn) ^Lqn. Thus
(4.3)
qn-
means.
qi
nqn
Hence
1
—>
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.
5. 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
\
1 —22 bnbu■• ' bin= 22 ( ?i^«'i+ 1ibi2+ • • • —qnbin+ II bikJ.
t-i
t-i \
*=i /
We can apphr Theorem
1 to each of the expressions
4 Loc. cit. §2.7, proof (ii) of Holder's inequality.
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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'</csn
required.
/
Clearly
it implies
that
E^?i^i2 ■ • • b"^—1 if, and only if, for each i, j and k the numbers
ba and ba are equal.
Henceforward we exclude this. In special cases certainly equality in
(5.1) can be attained.
For m = 2, m~^2, see §1. When m = n and
qi = q2= • ■ • =qn=l/n,
for instance,
there is equality
on the left
if and only if, for some permutation
k, A, • • • , w of the numbers
1, 2, • • • , n, bu = b2\= ■ ■ ■ =bmu = l; equality on the right if and
only if bu = b2\= • ■ ■ =bma = 0 while all the other bik are positive
and equal (i.e., = l/(m —1)).
6. A general, yet preliminary,
Theorem
3. Let m^2,
= 2-ii-i aik~>Q,
m
/
n
S'=E(E«.*),
result on Minkowski's
m^2,
r>l,
inequality.
r' = r(r —l)-1,
aik^0,
pt
\ r
Then
(6.1)
-~-r,22pkLk^22Pk-S^——~22p*Lk.
max (r, r ) k=i
k=i
mm (r, r ) k=i
Clearly there is equality whenever r = 2; and /"=1 pk>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
a«>
i~l
< E a'l°'i + E <*&?*+ • • • + E ain<*i\ ,
\ i=l
^
argument.6
i-1
r_1
hlr(
i-1
V/r'
We fix k ( = 1, or 2, • • • , n) and use Theorem
6 Loc. cit., §2.11, first proof of Minkowski's
/
inequality.
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2. If r>r'
let qi = l/r,
1958]
ARITHMETIC AND GEOMETRIC MEANS
457
52= 1/?"'; si= 22?-i Oik=pl, s2= 22T-i <r'i= Sr; we deduce that
(6.4)
PbST\l - qtLk) g 22 h\k(^)l'r' g nf\l
i=i
- qiLk).
If r<r' interchange qi and q2, and Si and s2 also. Combining these
results we can complete the proof. A theorem which is more suitable
for applications will be given in another paper.
Appendix. Added on October 10, 1957. We prove the Remark (b)
to Theorem 1; and in addition, two results: if n = 2, qi*tq2, the fraction in (1.2) tends to a unique limit when (xi —x0)2 + (x2 —xQ)2—>0,
where x0>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<qn.
If x2 = x3=
■ ■ ■ =xn = 0, Xi = x0 + e (e^O),
An(x)/(Un(x) -> 2(« - l)-1^^!
- <?i)
(e -» 0)
by the binomial series. If, however, Xi= ■ ■ ■ =x„_i = 0, x„ = xo + e,
the limit is 2(n —l)~1qn(l —q„) >2(n —l)~lqx(l —qi). Thus there is no
unique limit.
II. Let «>2,
qi = q2= ■ ■ ■ =qn = l/n.
existence of a unique limit. Then
Set Xi= Xf (i=l,
(1.2) un(x2n) =
22
{xt - Xk)2pik;
small e>0,
0<n2A2n-2-e<Pik<n2A2n-2
pik = ( £
+ f,
nAn(X2n)=22ixtn-nX21Xl
xT^xi)
■
there is a 5 such that
Now it follows from the Hurwitz-Muirhead
nomial
is 2n~2.
2, • • • , »), Xo= A2n. Then
Hence, given a sufficiently
(1.3)
We need only show the
(1.1) implies that the latter
j £
(X{ -
identity6
A)2 < b\ .
that
■ ■ ■ Xn is representable
the polyin the
form
(1.4)
wA„(Z2«)=
22
(Xi-XkYQik,
l£i<k$n
where the poynomials Qik = Qki are of degree 2n —2 and are homogeneous in the variables Xu X2, • ■ • , Xn, with all their coefficients
positive. Let K be the sum of the coefficients of Qik. Since Qjt may
• Loc. cit, §2.23 (2.23.1)and §2.21(1).
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458
H. KOBER
be derived from Qik by interchanging
K is independent
(1.5)
[June
Xi with Xt and Xk with Xt,
of i, k. Thus, for some 5>0,
0 < KA2"-2 - e < Qu, < KA1"-2 + t,
\}2(Xi-
A)2<&\
.
By (1.4) and (1.5),
A^2""2
-<-<-
-
e
nAn
n2A2n~2 + i
Un
KA2n~2 -+- i
n2A2n~2
-
€
As e—>0,A„/c7„—>Am~3; which shows the existence
III.
of the limit.
The case m = 2. We show that the function
/(*, y; P) = (px + qy-
x"y)/(xw
- y1'2)2
(0 <p
<p
+ q=
1)
tends to 2pq as (x —a)2-\-(y—a)2-+0,
where a>0 is fixed. If p is
rational, this follows readily7 from the preceding result.
Let now p be irrational
and let {Pm} and {pm} (m = l, 2, ■ ■ ■ ) be
sequences of rational numbers tending to p and such that Pm>p>pm.
Then
(1.6)
Qm—>g, qm—>q (w—><*>), where
lim
f(x,y;Pm)
Qm = l— Pm, qm=l—pm,
= 2PmQm;
lim
(i,l/)->(o,a)
and
f(x,y; pm) = 2pmqm.
(x,v)->(.a,a)
(i) Let x>y>0,
x/y = u (>1),
and consider the function
= sx-\-(l — s)y —x'yl~' for fixed x, y. Suppose
first that p<l/2.
may take Pm<l/2
g'(s)
-=
y
(m = l, 2, ■ ■ ■) and s<l/2.
g(s)
We
Then 2pmqm<2pq
u — 1 — us log u > u — 1 — u1'2 log u
/'«
i
(j,l/4
_
B-l/4)2_
dv
2v
> 0.
Hence g(P,,)>g(P)>g(p,);
(1.7)
2pmqmS
as g(J>ro)= (x"2-y^)2f(pm),
lim inf f(x, y; p) g
(x,v)-»(a,a)
Let now p>l/2.
Then
Ol/2,c<pm<p<Pm<l
Then w0> 1,
by (1.6)
lim sup f(x, y; p) g 2PmQm.
(i,K)->(a,o)
we may suppose that, for some constant
(m = l,2, ■ ■ •).SetM0={c(l-c)-1}1/(1-c).
i Loc. cit., cf. §2.2 and the proof of (2.5.1).
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1958I
ARTITHMETIC AND GEOMETRIC MEANS
g'(s)
u - 1 - uc log u
yu"
uc
-^-=
/'"
/•»
I
J i
dv rv(
1
1 V^+'Ji \uf°
459
{(1 - c)v + c - v°}-
dv
v1+c
1\
~ 2^/
2'
which is certainly negative for l<u<u0.
Since x—>a and y-^a, we
may take u, i.e. x/y, <u0. Hence/(x,
y; Pm) <f(x, y; p) <f(x, y; pm),
and we obtain an inequality
similar to (1.7), with
interchanged.
Taking m—><x>,we have
(1.8)
lim
f(x,y,p)
2pmqm and 2PmQm
= 2pq.
(x,v)->(a,a)
(ii) Let now y>x. We arrive at the same result when observing
that/(x,
y; p) =/(y, x; q)^>2qp; thus we can complete the proof.
IV. The last statement is deduced by an argument
used in I. Thus all the assertions are true.
Birmingham,
England
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similar to that