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arXiv:1305.3403v1 [math.CA] 15 May 2013
ON A MIXED ARITHMETIC-GEOMETRIC MEAN INEQUALITY
PENG GAO
Abstract. We extend a result of Holland on a mixed arithmetic-geometric mean inequality.
1. Introduction
Let Mn,r(q,x) be the generalized weighted power means: Mn,r(q,x) = (Pn
i=1 qixr
i)1
r, where
q= (q1, q2,··· , qn), x= (x1, x2,··· , xn), qi>0,1≤i≤nwith Pn
i=1 qi= 1. Here Mn,0(q,x)
denotes the limit of Mn,r (q,x) as r→0+. Unless specified, we always assume xi>0,1≤i≤n.
When there is no risk of confusion, we shall write Mn,r for Mn,r(q,x) and we also denote An, Gn
for the arithmetic mean Mn,1, geometric mean Mn,0, respectively.
For fixed x= (x1,··· , xn),w= (w1,··· , wn) with w1>0, wi≥0, we define xi= (x1,··· , xi),
wi= (w1,··· , wi), Wi=Pi
j=1 wj,Mi,r =Mi,r(wi/Wi,xi),Mi,r = (M1,r,··· , Mi,r). The following
result on mixed mean inequalities is due to Nanjundiah [5] (see also [1]):
Theorem 1.1. Let r > s and n≥2. If for 2≤k≤n−1,Wnwk−Wkwn>0. Then
Mn,s(Mn,r )≥Mn,r (Mn,s),
with equality holding if and only if x1=···=xn.
It is easy to see that the case r= 1, s = 0 of Theorem 1.1 follows from the following Popoviciu-
type inequalities established in [4] (see also [1, Theorem 9]):
Theorem 1.2. Let n≥2. If for 2≤k≤n−1,Wnwk−Wkwn>0, then
Wn−1ln Mn−1,0(Mn−1,1)−ln Mn−1,1(Mn−1,0)≤Wnln Mn,0(Mn,1)−ln Mn,1(Mn,0)
with equality holding if and only if xn=Mn−1,0=Mn−1,1(Mn−1,0).
In [6], the following Rado-type inequalities were established:
Theorem 1.3. Let s < 1and n≥2. If for 2≤k≤n−1,Wnwk−Wkwn>0, then
Wn−1Mn−1,s(Mn−1,1)−Mn−1,1(Mn−1,s)≤WnMn,s(Mn,1)−Mn,1(Mn,s )
with equality holding if and only if x1=···=xnand the above inequality reverses when s > 1.
The above theorem is readily seen to imply Theorem 1.1. In [3], Holland further improved the
condition in Theorem 1.3 for the case s= 0 by proving the following:
Theorem 1.4. Let n≥2. If W2
n−1≥wnPn−2
i=1 Wiwith the empty sum being 0, then
(1.1) Wn−1Mn−1,0(Mn−1,1)−Mn−1,1(Mn−1,0)≤WnMn,0(Mn,1)−Mn,1(Mn,0)
with equality holding if and only if x1=···=xn.
It is our goal in this paper to extend the above result of Holland by considering the validity of
inequality (1.1) for the case W2
n−1< wnPn−2
i=1 Wi. Note that this only happens when n≥3. In the
next section, we apply the approach in [2] to prove the following
2000 Mathematics Subject Classification. Primary 26D15.
Key words and phrases. Arithmetic-geometric mean, mixed-mean inequality.
1
2 PENG GAO
Theorem 1.5. Let n≥3. Inequality (1.1) holds when the following conditions are satisfied:
0<wnPn−2
i=1 Wi
W2
n−1
−1≤w1
wn
,Wn−1
Wn
n−2
Y
i=1 Wi+1
Wi
Wiwn
W2
n−1≤1,(1.2)
Wn−1wn
Wnw1 n−2
X
i=1
Wiwn
W2
n−1
−1!+wn
Wn!n−2
Y
i=1 Wi+1
wi+1
wi+1
Wn−1≤1.
In the above theorem, we do not give the condition for the equality in (1.1) to hold. In Section
3, we show that there do exist sequences {wi}n
i=1 that satisfy the conditions of Theorem 1.5.
2. Proof of Theorem 1.5
We may assume that xi>0, wi>0,1≤i≤nand the case xi= 0 or wi= 0 for some ifollows
by continuity. We recast (1.1) as
(2.1) Gn(An)−Wn−1
Wn
Gn−1(An−1)−wn
Wn
Gn≥0.
Note that
Gn(An) = Gn−1(An−1)Wn−1/Wn
Awn/Wn
n, Gn−1(An−1) = An
n−1
Y
i=1 Ai
Ai+1 Wi/Wn−1.(2.2)
Dividing Gn(An) on both sides of (2.1) and using (2.2), we can recast (2.1) as:
(2.3) Wn−1
Wn
n−1
Y
i=1 Ai
Ai+1 Wiwn/(Wn−1Wn)+wn
Wn
n
Y
i=1 xi
Aiwi/Wn
≤1.
We express xi= (WiAi−Wi−1Ai−1)/wi,1≤i≤nwith W0=A0= 0 to recast (2.3) as
Wn−1
Wn
n−1
Y
i=1 Ai
Ai+1 Wiwn/(Wn−1Wn)+wn
Wn
n
Y
i=1 WiAi−Wi−1Ai−1
wiAiwi/Wn
≤1.
We set yi=Ai/Ai+1, 1 ≤i≤2 to further recast the above inequality as
(2.4) Wn−1
Wn
n−1
Y
i=1
yWiwn/(Wn−1Wn)
i+wn
Wn
n−1
Y
i=1 Wi+1
wi+1
−Wi
wi+1
yiwi+1/Wn
≤1.
We now regard the right-hand side expression above as a function of yn−1only and define
f(yn−1) = Wn−1
Wn
c·ywn/Wn
n−1+wn
Wn
c′·Wn
wn
−Wn−1
wn
yn−1wn/Wn
,
where
c=
n−2
Y
i=1
yWiwn/(Wn−1Wn)
i, c′=
n−2
Y
i=1 Wi+1
wi+1
−Wi
wi+1
yiwi+1/Wn
.
On setting f′(yn−1) = 0, we find that
yn−1= Wn−1
Wn
+wn
Wnc′
cWn/Wn−1!−1
.
It is easy to see that f(yn−1) is maximized at the above value with its maximal value being
Wn−1
Wn
cWn/Wn−1+wn
Wn
(c′)Wn/Wn−1Wn−1/Wn
.
ON A MIXED ARITHMETIC-GEOMETRIC MEAN INEQUALITY 3
Thus, in order for inequality (2.4) to hold, it suffices to have
Wn−1
Wn
cWn/Wn−1+wn
Wn
(c′)Wn/Wn−1≤1.
Explicitly, the above inequality is
g(y1, y2,...,yn−2) := Wn−1
Wn
n−2
Y
i=1
yWiwn/W 2
n−1
i+wn
Wn
n−2
Y
i=1 Wi+1
wi+1
−Wi
wi+1
yiwi+1/Wn−1≤1.
Let (a1, a2,...,an−2)∈[0, W2/W1]×[0, W3/W2]×...×[0, Wn−1/Wn−2] be the point in which
the absolute maximum of gis reached. If one of the aiequals 0 or Wi+1 /Wi, then it is easy to see
that we have
g(a1, a2,...,an−2)≤max Wn−1
Wn
n−2
Y
i=1 Wi+1
WiWiwn/W 2
n−1
,wn
Wn
n−2
Y
i=1 Wi+1
wi+1 wi+1/Wn−1!.(2.5)
If the point (a1, a2,...,an−2) is an interior point, then we have
∇g(a1, a2,...,an−2) = 0.
It follows that for every 1 ≤i≤n−2, we have
Qn−2
i=1 aWiwn/W 2
n−1
i
Qn−2
i=1 Wi+1
wi+1 −Wi
wi+1 aiwi+1/Wn−1=ai
Wi+1
wi+1 −Wi
wi+1 ai
:= 1
d,(2.6)
where d > 0 is a constant (depending on the wi). In terms of d, we have
ai=Wi+1
dwi+1 +Wi
.
We use this to recast the first equation in (2.6) as
n−2
Y
i=1 Wi+1
dwi+1 +WiWiwn/W 2
n−1
=1
d
n−2
Y
i=1 Wi+1
wi+1
·dwi+1
dwi+1 +Wiwi+1/Wn−1.
We recast the above equality as
ln n−2
Y
i=1
(Wi+1)Wiwn/W 2
n−1−wi+1/Wn−1!
=
n−2
X
i=1 Wiwn
W2
n−1
−wi+1
Wn−1ln (dwi+1 +Wi)−w1
Wn−1
ln d:= h(d).
Note that the above equality holds when d= 1 and we have
h′(d) =
n−2
X
i=1 Wiwn
W2
n−1
−wi+1
Wn−11
d+Wi/wi+1
−w1
Wn−1
1
d.
As
Wiwn
W2
n−1
−wi+1
Wn−1
≥0⇔Wi
wi+1
≥Wn−1
wn
,
4 PENG GAO
it follows that
h′(d)≤
n−2
X
i=1 Wiwn
W2
n−1
−wi+1
Wn−11
d+Wn−1/wn
−w1
Wn−1
1
d
≤Pn−2
i=1 Wiwn
W2
n−1
−1d−w1
wn
d(d+Wn−1/wn).
Therefore, when
n−2
X
i=1
Wiwn
W2
n−1
−1≤0,
the function h(d) is a decreasing function of dso that d= 1 is the only value that satisfies (2.6) and
we have ai= 1 correspondingly with g(1,1,...,1) = 1 and this allows us to recover Theorem 1.4,
by combining the observation that the right-hand side expression of (2.5) is an increasing function
of wnfor fixed wi,1≤i≤n−1 with the discussion in the next section.
Suppose now
n−2
X
i=1
Wiwn
W2
n−1
−1>0,(2.7)
then the function h(d) is a decreasing function of dfor
d≤
w1
wn
Pn−2
i=1 Wiwn
W2
n−1
−1:= d0.
If follows that if d0≥1, then d= 1 is the only value ≤d0that satisfies (2.6) and we have ai= 1
correspondingly with g(1,1,...,1) = 1. We further note that for any d≥d0satisfying (2.6), the
value of gat the corresponding aisatisfies
g(a1, a2,...,an−2) = Wn−1
dWn
+wn
Wnn−2
Y
i=1 Wi+1
wi+1
·dwi+1
dwi+1 +Wiwi+1/Wn−1
≤Wn−1
dWn
+wn
Wnn−2
Y
i=1 Wi+1
wi+1 wi+1/Wn−1
≤Wn−1
d0Wn
+wn
Wnn−2
Y
i=1 Wi+1
wi+1 wi+1/Wn−1.
Combining this with (2.5), we see that inequality (2.4) holds when the conditions in (1.2) are
satisfied and this completes the proof of Theorem 1.5.
3. A Further Discussion
We show in this section that there does exist sequences {wi}n
i=1 satisfying the conditions of
Theorem 1.5. To see this, we note that the left-hand side expression of (2.7) vanishes when
wn=W2
n−1
Pn−2
i=1 Wi
.(3.1)
It follows by continuity that such sequences {wi}n
i=1 satisfying the conditions of Theorem 1.5
exist as long as the positive sequence {wi}n
i=1 with wi,1≤i≤n−1 being arbitrary and wndefined
ON A MIXED ARITHMETIC-GEOMETRIC MEAN INEQUALITY 5
by (3.1) satisfies the last two inequalities of (1.2) with strict inequalities there. It is readily checked
that these inequalities become
Pn−2
i=1 Wi
Pn−1
i=1 Wi!Pn−2
i=1 Wi
WWn−2
n−1<
n−2
Y
i=1
Wwi
i,(3.2)
Wn−1
Pn−1
i=1 Wi
n−2
Y
i=1 Wi+1
wi+1 wi+1/Wn−1<1.(3.3)
Now, it is easy to see that inequality (3.2) holds when n= 3. It follows that it holds for all n≥3
by induction as long as we have
Pn−2
i=1 Wi
Pn−1
i=1 Wi!Pn−2
i=1 Wi
WWn−1
n−1≥ Pn−1
i=1 Wi
Pn
i=1 Wi!Pn−1
i=1 Wi
WWn−1
n.(3.4)
The right-hand side expression above when regarded as a function of Wnonly is maximized at
Wn=Wn−1Pn−1
i=1 Wi
Pn−2
i=1 Wi
.
As the inequality in (3.4) becomes an equality with this value of Wn, we see that inequality (3.2)
does hold for all n≥3.
Note that it follows from the arithmetic-geometric mean inequality that
Wn−1
Pn−1
i=1 Wi
n−2
Y
i=1 Wi+1
wi+1 wi+1/Wn−1≤Wn−1
Pn−1
i=1 Wi n−2
X
i=0
Wi+1
wi+1
·wi+1
Wn−1!= 1.
As one checks easily that the above inequality is strict in our case, we see that inequality (3.3)
also holds. We therefore conclude the existence of sequences {wi}n
i=1 satisfying the conditions of
Theorem 1.5.
References
[1] P. S. Bullen, Inequalities due to T. S. Nanjundiah, Recent progress in inequalities, Kluwer Acad. Publ., Dordrecht,
1998, 203-211.
[2] P. Gao, A note on mixed-mean inequalities, J. Inequal. Appl.,2010 (2010), Art. ID 509323, 8 pp.
[3] F. Holland, An inequality between compositions of weighted arithmetic and geometric means, JIPAM. J. Inequal.
Pure Appl. Math.,7(2006), Article 159, 8 pp. (electronic).
[4] K. Kedlaya, A weighted mixed-mean inequality, Amer. Math. Monthly,106 (1999), 355-358.
[5] T. S. Nanjundiah, Sharpening of some classical inequalities, Math Student,20 (1952), 24-25.
[6] C. Tarnavas and D. Tarnavas, An inequality for mixed power means, Math. Inequal. Appl.,2(1999), 175–181.
Department of Mathematics, School of Mathematics and System Sciences, Beijing University of
Aeronautics and Astronautics, P. R. China
E-mail address:penggao@buaa.edu.cn
