Exam (elaborations)
On Jensen Type Inequalities with Ordered Variables
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- Modular_Arithmetic_in_the_AMC_and_AIME.
On Jensen Type Inequalities with Ordered Variables
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Volume 9 (2008), Issue 1, Article 19, 14 pp.ON JENSEN TYPE INEQUALITIES WITH ORDERED VARIABLES
VASILE CÎRTOAJE
D EPARTMENT OF AUTOMATION AND C OMPUTERS
U NIVERSITY OF P LOIE ŞTI
B UCURE ŞTI 39, ROMANIA
vcirtoaje@upg-ploiesti.ro
Received 15 October, 2007; accepted 05 January, 2008
Communicated by J.E. Pecarić
A BSTRACT. In this paper, we present some basic results concerning an extension of Jensen
type inequalities with ordered variables to functions with inflection points, and then give several
relevant applications of these results.
Key words and phrases: Convex function, k-arithmetic ordered variables, k-geometric ordered variables, Jensen’s inequality,
Karamata’s inequality.
2000 Mathematics Subject Classification. 26D10, 26D20.
1. BASIC R ESULTS
An n-tuple of real numbers X = (x1 , x2 , . . . , xn ) is said to be increasingly ordered if x1 ≤
x2 ≤ · · · ≤ xn . If x1 ≥ x2 ≥ · · · ≥ xn , then X is decreasingly ordered.
In addition, a set X = (x1 , x2 , . . . , xn ) with x1 +x2 +···+x
n
n
= s is said to be k-arithmetic
ordered if k of the numbers x1 , x2 , . . . , xn are smaller than or equal to s, and the other n − k
are greater than or equal to s. On the assumption that x1 ≤ x2 ≤ · · · ≤ xn , X is k-arithmetic
ordered if
x1 ≤ · · · ≤ xk ≤ s ≤ xk+1 ≤ · · · ≤ xn .
It is easily seen that
X1 = (s − x1 + xk+1 , s − x2 + xk+2 , . . . , s − xn + xk )
is a k-arithmetic ordered set if X is increasingly ordered, and is an (n − k)-arithmetic ordered
set if X is decreasingly ordered.
√
Similarly, an n-tuple of positive real numbers A = (a1 , a2 , . . . , an ) with n a1 a2 · · · an = r is
said to be k-geometric ordered if k of the numbers a1 , a2 , . . . , an are smaller than or equal to r,
and the other n − k are greater than or equal to r. Notice that
ak+1 ak+2 ak
A1 = , ,...,
a1 a2 an
is a k-geometric ordered set if A is increasingly ordered, and is an (n − k)-geometric ordered
set if A is decreasingly ordered.
316-07
, 2 VASILE C ÎRTOAJE
Theorem 1.1. Let n ≥ 2 and 1 ≤ k ≤ n − 1 be natural numbers, and let f (u) be a function on
a real interval I, which is convex for u ≥ s, s ∈ I, and satisfies
f (x) + kf (y) ≥ (1 + k)f (s)
for any x, y ∈ I such that x ≤ y and x + ky = (1 + k)s. If x1 , x2 , . . . , xn ∈ I such that
x1 + x2 + · · · + xn
=S≥s
n
and at least n − k of x1 , x2 , . . . , xn are smaller than or equal to S, then
f (x1 ) + f (x2 ) + · · · + f (xn ) ≥ nf (S).
Proof. We will consider two cases: S = s and S > s.
A. Case S = s. Without loss of generality, assume that x1 ≤ x2 ≤ · · · ≤ xn . Since x1 + x2 +
· · · + xn = ns, and at least n − k of the numbers x1 , x2 , . . . , xn are smaller than or equal to s,
there exists an integer n − k ≤ i ≤ n − 1 such that (x1 , x2 , . . . , xn ) is an i-arithmetic ordered
set, i.e.
x1 ≤ · · · ≤ xi ≤ s ≤ xi+1 ≤ · · · ≤ xn .
By Jensen’s inequality for convex functions,
f (xi+1 ) + f (xi+2 ) + · · · + f (xn ) ≥ (n − i)f (z),
where
xi+1 + xi+2 + · · · + xn
z= , z ≥ s, z ∈ I.
n−i
Thus, it suffices to prove that
f (x1 ) + · · · + f (xi ) + (n − i)f (z) ≥ nf (s).
Let y1 , y2 , . . . , yi ∈ I be defined by
x1 + ky1 = (1 + k)s, x2 + ky2 = (1 + k)s, . . . , xi + kyi = (1 + k)s.
We will show that z ≥ y1 ≥ y2 ≥ · · · ≥ yi ≥ s. Indeed, we have
y1 ≥ y2 ≥ · · · ≥ yi ,
s − xi
yi − s = ≥ 0,
k
and
ky1 = (1 + k)s − x1
= (1 + k − n)s + x2 + · · · + xn
≤ (k + i − n)s + xi+1 + · · · + xn
= (k + i − n)s + (n − i)z ≤ kz.
Since z ≥ y1 ≥ y2 ≥ · · · ≥ yi ≥ s implies y1 , y2 , . . . , yi ∈ I, by hypothesis we have
f (x1 ) + kf (y1 ) ≥ (1 + k)f (s),
f (x2 ) + kf (y2 ) ≥ (1 + k)f (s),
.........
f (xi ) + kf (yi ) ≥ (1 + k)f (s).
Adding all these inequalities, we get
f (x1 ) + f (x2 ) + · · · + f (xi ) + k[f (y1 ) + f (y2 ) + · · · + f (yi )] ≥ i(1 + k)f (s).
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 19, 14 pp. http://jipam.vu.edu.au/
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