THE TEACHING OF MATHEMATICS
2007, Vol. X, 2, pp. 107–108
A PROOF OF CHEBYSHEV’S INEQUALITY
Milosav M. Marjanović and Zoran Kadelburg
Abstract. Interrelating inequalities by proving that one of them is a specific
case of others, makes their proofs transparent and often easier. Thus, we derive here
Chebyshev’s inequality from two inequalities related to convex combinations (and also
having some interest in themselves).
ZDM Subject Classification: I34; AMS Subject Classification: 00A35.
Key words and phrases: Relation of majorization, convex combinations, Chebyshev’s inequality.
This short note is just an addendum to the paper [1].
Recall that, for decreasing sequences a = (αi )ni=1 and b = (βi )ni=1 of real
numbers, a is said to majorize b, what is denoted by a  b, if the terms of these
sequences satisfy the following two conditions:
1◦ α1 + α2 + · · · + αi > β1 + β2 + · · · + βi , for each i ∈ {1, 2, . . . , n − 1};
2◦ α1 + α2 + · · · + αn = β1 + β2 + · · · + βn .
First, we prove the following two lemmas.
Lemma 1. Let a = (αi )ni=1 , b = (βi )ni=1 and x = (xi )ni=1 be three decreasing
sequences of real numbers, such that a  b. Then the following inequality holds:
n
n
X
X
αi xi >
βi xi .
i=1
i
P
i=1
i
P
Proof. Denote Ai =
αj , Bi =
βj , for i ∈ {1, 2, . . . , n}, and put A0 =
j=1
B0 = 0. Then we have j=1
n
n
n
n
X
X
X
X
αi xi −
βi xi =
(αi − βi )xi =
(Ai − Ai−1 − Bi + Bi−1 )xi
i=1
i=1
i=1
=
=
n
X
(Ai − Bi )xi −
(Ai−1 − Bi−1 )xi
i=1
i=1
n−1
X
n−1
X
i=1
=
i=1
n
X
n−1
X
(Ai − Bi )xi −
(Ai − Bi )xi+1
i=0
(Ai − Bi )(xi − xi+1 ) > 0,
i=1
being Ai − Bi > 0 and xi − xi+1 > 0 for each i ∈ {1, 2, . . . , n − 1}.
108
M. M. Marjanović, Z. Kadelburg
In particular, when a and b are decreasing, a  b and
above inequality holds for convex combinations of points
n
P
αi =
i=1
x1 , x2 ,
n
P
i=1
βi = 1, the
. . . , xn .
Lemma 2. Let a = (αi )ni=1 be a decreasing sequence of nonnegative numbers
n
P
1
such that
αi = 1 and let b = (βi )ni=1 with βi = . Then a  b.
i
i=1
1
Proof. It is clear that α1 > , and let us suppose that
n
i
α1 + · · · + αi > ,
n
i+1
, what would immediatefor some i. It is not true that α1 + · · · + αi+1 <
n
n
P
1
ly imply αi+1 < . From
αi = 1 it follows that αi+2 + · · · + αn = 1 −
n
i=1
n − (i + 1)
1
(α1 + · · · + αi+1 ) >
, and so αi+2 > , which would contradict the fact
n
n
i+1
that a is decreasing. Hence, α1 + · · · + αi+1 >
and the inductive proof that
n
i
i
P
P
αj >
βj for each i ∈ {1, 2, . . . , n − 1} is completed.
j=1
j=1
Now, we derive the well-known Chebyshev’s inequality
µX
¶µX
¶
n
n
n
X
ai
bi 6 n
ai bi .
i=1
i=1
i=1
for decreasing sequences (ai )ni=1 , (bi )ni=1 of real numbers.
Proof. Without loss of generality, we can assume that the terms ai and bi
of the given sequences are nonnegative (if, for example, some of ai ’s or bi ’s were
negative, we would apply the procedure that follows to the terms a0i = ai − an > 0
and b0i = bi − bn > 0).
n
P
1
ai
and βi = . Then the sequences (αi ) i (βi ) are
Denote A =
ai , αi =
A
n
i=1
decreasing and, by Lemma 2, (αi ) Â (βi ) holds true. Applying Lemma 1 to the
sequences (αi ), (βi ) and taking xi = bi , we obtain
n
n
X
X
ai
1
· bi >
· bi ,
A
n
i=1
i=1
µ n ¶µ n ¶
n
P
P
P
i.e., n
ai bi >
ai
bi .
i=1
i=1
i=1
REFERENCES
- ukić, M. Lukić, I. Matić: Inequalities of Karamata, Schur and Muirhead,
[1] Z. Kadelburg, D. D
and some applications, The Teaching of Mathematics, Vol. VIII, 1 (2005), 31–45.
Milosav M. Marjanović, Mathematical Institute SANU, Kneza Mihaila 35/I, 11000 Belgrade,
Serbia, E-mail: [email protected]
Zoran Kadelburg, Faculty of Mathematics, Studentski trg 16/IV, 11000 Belgrade, Serbia,
E-mail: [email protected]