I nternat. J. 4ath. & Math. Si.
531
Vol. 2 #3 (1979) 531-535
AN ELEMENTARY INEQUALITY
SEYMOUR HABER
United States Department of Commerce
National Bureau of Standards
Washington, D.C.
20234
(Received November 8, 1978)
An elementary inequality is proved in this note.
ABSTRACT.
KEY WORDS AND PHRASES.
Inequy, Compound interest.
AMS (MOS) SUBJECT CLASSIFICATION (1970) CODES.
I.
26DT5.
INTRODUCTION AND RESULTS.
The following theorems present a series of closely related inequalities.
form that was found originally is given first.
THEOREM i.
a, b
If
n
a
PROOF.
+
_> 0
an-I
be defined by
0
b
n
is a positive integer, then
+ an-2 b 2 +’’’+
n+ I
b
n
>
(a +2
Let Ix] be, as usual, the integer part of
[n/2 ],
symbol
and
b
n
x, and let the
(I i)
The
532
S. HABER
x
Enl2]
o
x
0
+
0
+ x I +...+ X[n/2]_ I +
x
+...+ X[n/2 ]
1
x
1
if
x[n/2]
n
is even.
n
a > b, we divide through by a in
Assuming, without loss of generality, that
and set
is odd
n
if
(I.I),
b/a, obtaining
x+ x 2 +...+ xn
1+
(1 +2 x)
>
n+l
n
O-<x<- 1
(1.2)
Now
(i
En/2]
x)n
+
*
o
i
()
(x +
xn-f)
and
i
+
x
+ x 2 +...+ xn
i
<x +
xn-i
0
so that we may rewrite (1.2) as
()) (x i + xn-i) >- 0
2n
I + x
We now note that
> x
+ xn-I
> x
2
+ xn-2 >...>
an example of the rearrangements inequality
k
X
+X
n-k
k+I/2
X
X
0 < x -< I
x
[n/2]
(1.3)
+ xn-[n/2]
([I ], p.261), since we can
-I/2
n-k-I/2
+X
This is
write
XI/2
and
X
and for
and x
i/2
k+l
0 < k _< [n/2]
+ Xn-k-I
k+I/2
I, x
It follows that if we set
X
k+I/2
and
X
1/2
+ Xn-k-I/2 X -1/2
n-k-i/2
x
are in the same order as x
-1/2
533
ELEMENTARY INEQUALITY
n-i
if i < n/2
xi+x
a
i
x
I
n-i
+
x
then
a
0
->
a
>
>
I
when
n+l
b >-...>
I -> 2
2
(n)
i
n
In/2]
I, 2,
i
and since (i.i)
b[ n/2]
b.
b, the sum of the
a
n/2
is even and i
If we also set
a[n/2 ].
bi
it is clear that b
n
if
2
(1.3) are equalities
(1.3) is then an immediate consequence of
is zero.
the following obvious lemma, which is also related to the rearrangements inequality.
LEMMA.
b
I
> b
2
k
If
>...> b
is a positive integer and a
b
and
k
I
+
b
2
+...+
b
I
0, then
k
> a
2
>...> a
k
alb I + a2b 2
>_
0, and
+...+
akbk
> 0.
COROLLARIES.
n
x -I > n(x-l)
n
x-I
n(x-l)
<
)n-I
x+l.
x >_ 1
and
(1.4)
0,1,2,
n
n-I
0 _< x _< I
(--)
and
0,1,2,
n
(1.5)
Inequality (1.4) is a sharpening of Hardy, Littlewood, and Polya’s (2.15.3)
for part of the latter’s range, while (1.5) is complementary to (2.15.3) for another
These are immediate consequences of (1.2), which is valid for
part of its range.
all
x _> 0
by Theorem i.
(The case
(1.2) holds for
is obvious; similarly
is not a consequence of Theorem I. but
0
n
n
0
if we interpret
I + x + x 2 +...+ x
n
as 1 for that case.)
Setting
(1.4) and (1.5) gives us the inequalities
n-I
n
)
t > 0 and n
(l+t) > l+nt(l+,
0,1,2,
x
l+t
in
(1.6)
and
(l+t)
Putting
-t
n
<
)
l+nt(l+
in place of
t
n-I
-I
_< t < 0
and
n
0,1,2
gives an alternative form:
(1.7)
S. HABER
534
n-I
Replacing
I/t
by
t
(t+l)
l-nt(l-)
<
(l-t)
n > n
t
in
0 < t -<
and
(1.8)
0,1,2,
n
(1.6) gives us
n-i
+ n(t+ )
This is better than (1.6) for
t > 1
t > 0
and
(1.9)
0,1,2,
n
though not as good for
(for n > 2).
t <
Similarly, from (1.8) we obtain
I n-I
< tn
(t_l)n
n(t-)
t >
and
(I.I0)
0,1,2,
n
(1.6) is a sharpening of the observation that compound interest beats simple
interest:
a period rate
n
periods, is better than simple
t(l+t/2) n-I
interest at the rate
(1.6)
t, compounded for
(i.i0) were obtained from (1.4) or (1.5) by invertible bilinear changes
of variable; similarly each of (1.4) and (1.5) may be obtained from the other by
replacing
x
by
I/x.
Thus (1.4)-(1.10) are equivalent, and equivalent to Theorem I.
Another bilinear transformation- setting
(l+t)
n
which is obvious’.
(l-t)
n >
2nt
x
0 < t <
(l+t)/(l-t)
and
in (1.4)
gives us
(I.II)
0,1,2,
n
This provides a quick alternate proof of Theorem i., but one
which does not show the donnection with the rearrangements inequality.
The form of inequalities (1.4)
n.
(i. ii) suggests consideration of non-integral
The inequalities are equivalent in this precise sense;
for any given
n, one of
these inequalities holds for its stated range of values of the other variable
(x or t) if and only if all the other inequalities hold, for that
stated ranges of the other variable.
Thus for each
n
n, for their
(1.4)
we may choose which of
(I.II) to study, at our convenience.
For
0 < n < I
(and 0 < t < I)
the binomial expansion of (l+t)
n
(l-t)
n
is
ELEMENTARY INEQUALITY
() ()
and each of the coefficients
holds for
For
between
n
n
0
I
between
535
is positive.
It follows that
I.
I.
and
2, all of
and
(), (),
are negative.
So for
I < n < 2 (and in fact I < n < 2), (I.II) holds with the inequality reversed.
For non-integral
n > 2
we look at (1.6).
i
+
f (t,n)
nt(l) n-I
(l+t)
f(0,n)
is
< 0
I.
-
Set
n
To show that (1.6) holds, it is sufficient to show that
for
(I +
(l+t)n+l
t > 0, n > 2.
Setting
( -) (
This is certainly negative when
)
I
n-2, the quantity in brackets is
s
.
t s
+)
st/2
1
(I +
is negative; when the latter is non-
negative, the whole quantity is
st
<
(z---)
(e
z_<o
Summing up we have
THEOREM 2.
Each of the inequalities (1.4)
(I.II) holds also for non-lntegral
n, when 0 < n < I and when n > 2: for I < n < 2 each of (1.4)
(i.ii) holds with
the direction of the inequality reversed.
REFERENCES
i.
Hardy, G.H., J.E. Littlewood and G. Polya, Inequalities, Cambridge University
Press
(1959).