THE TEACHING OF MATHEMATICS
2013, Vol. XVI, 1, pp. 1–5
GENERALIZATIONS OF SOME REMARKABLE INEQUALITIES
D.M. Bǎtineţu-Giurgiu, Neculai Stanciu
Abstract. In this paper, we present a generalization of Mitrinović’s inequality for
polygons, and triangles, a generalization of J. Radon’s inequality and a generalization
of Nesbitt’s inequality. The main tool in the proofs is the inequality of Jensen.
MathEduc Subject Classification: H34
MSC Subject Classification: 97H30
Key words and phrases: Jensen inequality; Mitrinović inequality; J. Radon inequality; Bergström inequality; Nesbitt inequality.
1. Generalizations of Mitrinović’s inequality
for convex polygons and triangles
If A1 A2 . . . An , n > 3 is a convex polygon, and M ∈ int(A1 A2 . . . An ), with
prAk Ak+1 M = Tk ∈ [Ak Ak+1 ] for k ∈ {1, 2, . . . , n}, An+1 = A1 , then
n
X
π
Ak Ak+1
> 2n tan .
M Tk
n
k=1
Proof. We first prove the following
Lemma. Let A, B, A 6= B be points in the plane and M ∈
/ AB, T = prAB M .
Then
AB
= tan u + tan v,
MT
where u = µ(∠AM T ), v = µ(∠T M B) are the measures of the angles ∠AM T and
∠T M B.
Proof of the Lemma. We have the following cases:
AT
BT
AB
i) T ∈ (AB). Then tan u =
and tan v =
, hence tan u + tan v =
.
MT
MT
MT
AT
AA
BT
ii) T ≡ A. Then tan u =
=
= 0 and tan v =
, hence tan u +
MT
MT
MT
AB
tan v =
. Similarly if T ≡ B.
MT
Ak Ak+1
From Lemma, we have
= tan uk + tan vk , for k = 1, n, where uk =
M Tk
µ(∠Ak M Tk ), vk = µ(∠Tk M Ak+1 ). It follows that
n
n
X
X
Ak Ak+1
=
(tan uk + tan vk ).
M Tk
k=1
k=1
2
D.M. Bǎtineţu-Giurgiu, N. Stanciu
Since the function f : [0, π/2) → [0, +∞), f (x) = tan x is convex on [0, π/2), we
can apply Jensen’s inequality and obtain that
¶
µ
n
n
X
Ak Ak+1
1 X
(uk + vk ) .
> 2n tan
M Tk
2n
k=1
k=1
Pn
Since, k=1 (uk + vk ) = 2π, we deduce that
n
X
Ak Ak+1
k=1
M Tk
> 2n tan
2π
π
= 2n tan ,
2n
n
and we are done.
Remark 1.1.
M ≡ I, we have
1 Pn
Ak Ak+1 =
r k=1
If A1 A2 . . . An is circumscribed about the circle C(I; r) and
M Tk = r, k = 1, n, and the obtained inequality becomes
2s
π
> 2n tan , wherefrom,
r
n
π
(1)
s > nr tan .
n
The inequality (1) is a generalization of Mitrinović’s inequality for triangles
√
(M)
s > 3r 3.
Remark 1.2. If A1 A2 A3 is a triangle, then the obtained inequality becomes
√
A1 A2
A2 A3
A3 A1
π
+
+
> 6 tan = 6 3.
M T1
M T2
M T3
3
For M ≡ I, we obtain (M). For more results see [1].
2. A generalization of J. Radon’s inequality
In what follows, we denote R+ = [0, +∞) and R∗+ = (0, +∞).
Pn
Pn
Let a, b, c, d, xk , yk ∈ R∗+ , k = 1, n and Xn =
k=1 . If
k=1 xk , Yn =
s
s
m, p, q, s ∈ R+ , r ∈ [1, +∞) are such that cYn > d max16k6n yk , k = 1, n, then
(2)
n
r(m+1)
X
(aX p + bxq )m+1 x
n
k=1
k
(cYns
−
k
dyks )m ykm
>
(anq Xnp+r + bXnq+r )m+1
m(s+1)
d)m Yn
(cns −
·
1
n(m+1)(q+r−1)−ms
Proof. Denote uk = (aXnp +bxqk )xrk , vk = (cYns −dyks )yk , k = 1, n, Vn =
and the left-hand side of (2) becomes
µ ¶m+1
µ ¶m+1
n
n
n
X
X
X
um+1
uk
vk uk
k
=
vk
= Vn
.
vkm
vk
Vn v k
k=1
k=1
R∗+
R∗+ ,
Pn
k=1
.
vk
k=1
m+1
Since the function f :
→
f (x) = x
is convex, we use Jensen’s inequality
and we obtain that
µ ¶
µX
¶
µX
¶
µX
¶m+1
n
n
n
n
X
uk
vk u k
uk
1
vk
f
>f
·
=f
= m+1
uk
.
Vn
vk
Vn v k
Vn
Vn
k=1
k=1
k=1
k=1
3
Generalizations of some remarkable inequalities
Therefore,
n
X
um+1
k
k=1
vkm
>
µX
n
Vn
Vnm+1
¶m+1
uk
k=1
i.e.,
k=1
Ã
n
r(m+1)
X
(aX p + bxq )m+1 x
n
k
(cYns
k=1
³
=
−
k
dyks )m ykm
n
P
k=1
>
(aXnp + bxqk )xrk
³P
n
aXnp
k=1
xrk + b
n
P
k=1
xq+r
k
´m+1
k=1
´m
³
´m+1
n
n
P
P
aXnp
xrk + b
xq+r
k
=
³
´m
n
n
P
P
cYns
yk − d
yks+1
´m+1
(cYns − dyks )yk
k=1
n
P
µ n
¶m+1
1 X
= m
uk
,
Vn
k=1
k=1
³
´m
n
P
cYns+1 − d
yks+1
.
k=1
k=1
Since the functions g, h, k : R∗+ → R∗+ , g(x) = xr , h(x) = xq+r , k(y) = y s+1 are
convex, also by Jensen’s inequality, we have:
µ X
¶
n
n
n
X
X
Xnr
Xr
1
r
xk =
xk = n · rn = r−1
,
g(xk ) > ng
n
n
n
k=1
k=1
k=1
µ X
¶
n
n
n
X
X
1
Xnq+r
Xnq+r
xq+r
=
h(x
)
>
nh
x
= n · q+r
= q+r−1
,
k
k
k
n
n
n
k=1
k=1
k=1
µ X
¶
n
n
n
X
X
1
Y s+1
Y s+1
s+1
yi =
k(yi ) > nk
yi = n · ns+1 = n s .
n i=1
n
n
i=1
i=1
Then, we deduce that
³
n
r(m+1)
X
(aX p + bxq )m+1 x
n
k
k
(cYns − dyks )m ykm
k=1
=
=
a·
>
(anq Xnp+r + bXnq+r )m+1
m(s+1)
(cns − d)m Yn
(anq Xnp+r + bXnq+r )m+1
(cns
m(s+1)
d)m Yn
·
·
Xnq+r ´m+1
Xnp+r
+
b
·
nr−1
nq+r−1
³
dY s+1 ´m
cYns+1 − ns
n
nms
m(m+1)(q+r−1)
1
,
n(m+1)(q+r−1)−ms
−
and we are done.
Remark 2.1. If p = q = s = 0 then (2) becomes
n
r(m+1)
X
(a + b)m+1 x
k
k=1
(c −
d)m ykm
r(m+1)
>
(a + b)m+1 Xn
(c − d)m Ynm
i.e.,
(20 )
n
r(m+1)
X
x
k
k=1
ykm
r(m+1)
>
Xn
.
Ynm n(m+1)(r−1)
·
1
n(m+1)(r−1)
,
4
D.M. Bǎtineţu-Giurgiu, N. Stanciu
If we consider r = 1 then by (20 ) we obtain
n
X
xm+1
Xnm+1
k
(R)
>
,
ykm
Ynm
k=1
i.e., just the inequality of J. Radon (see, e.g., [2]), with equality if and only if there
exists t ∈ R∗+ such that xk = tyk , k = 1, n.
Remark 2.2. If m = 1 then (2) becomes
n
X
(aXnp + bxqk )2 x2r
(anq Xnp+r + bXnq+r )2
1
k
(200 )
>
· 2(q+r−1)−s .
s+1
s
s
s
(cYn − dyk )yk
n
(cn − d)Yn
k=1
If we take p = q = s = 0, r = 1 then by (200 ) we obtain
n
X
X2
x2k
(B)
> n.
yk
Yn
k=1
But, that is just the inequality of H. Bergström.
3. A generalization of Nesbitt’s inequality
Pn
If a ∈ R+ , b, c, d, xk ∈ R∗+ , k = 1, n, Xn =
k=1 xk , m ∈ [1, +∞) and
,
then
cXnm > d max16k6n xm
k
(OG)
n
X
aXn + bxk
(an + b)nm 1−m
>
Xn .
m
cXnm − dxk
cnm − d
k=1
Proof. We have
n
n
X
X
(aXn + bxk )2
aXn + bxk
=
Un =
cXnm − dxm
(aXn + bxk )(cXnm − dxm
k
k )
k=1
k=1
n
X
(aXn + bxk )2
=
m+1 ,
m
acXnm+1 − adXn xm
k + bcXn xk − bdxk
k=1
where we apply H. Bergström inequality (B) and we deduce that
³P
´2
n
(aXn + bxk )
k=1
Un >
n
n
n
P
P
P
m+1
m
xk − bd
xm+1
acnXn
− adXn
xm
k + bcXn
k
k=1
k=1
k=1
(anXn + bXn )2
=
.
n
n
P
P
(acn + bc)Xnm+1 − adXn
xm
xm+1
k − bd
k
k=1
k=1
Using convexity of the functions f, g : R∗+ → R∗+ , f (x) = xm , g(x) = xm+1 , by
Jensen’s inequality we have:
n
n
X
X
Xnm
X m+1
xm
>
and
xm+1
> nm ,
k
k
m−1
n
n
k=1
k=1
Generalizations of some remarkable inequalities
5
and we obtain that
(an + b)2 Xn2
adXnm+1
bdXnm+1
(acn + bc)Xnm+1 −
−
nm−1
nm
(an + b)2 nm
X 1−m
=
acnm+1 + bcnm − adn − bd n
(an + b)2 nm
(an + b)nm 1−m
1−m
=
X
=
Xn ,
n
(an + b)(cnm − d)
cnm − d
which completes the proof.
Remark 3.1. If a = 0 and b = c = d = m = 1 then we obtain the Nesbitt’s
inequality for n variables, i.e.,
n
X
xk
n
(N)
.
>
Xn − xk
n−1
Un >
k=1
If we take n = 3 then by (N) we obtain
x1
x2
x3
3
+
+
> ,
x2 + x3
x3 + x1
x1 + x2
2
i.e., Problem 15114, proposed by A.M. Nesbitt to Educational Times 3 (1903),
37–38.
Remark 3.2. A generalization of (OG) was published in [3], i.e., if a, m ∈ R+ ,
n
P
b, c, d, xk ∈ R∗+ , k = 1, n, Xn =
xk , p ∈ [1, +∞), and cXnm > d max16k6n xm
k ,
k=1
then
n
X
(AMM)
k=1
aXn + bxk
(an + b)nmp 1−mp
.
>
X
p
(cXnm − dxm
(cnm − d)p n
k )
Remark 3.3. A generalization of (AMM) appeared in [4], i.e., if n ∈ N∗ \ {1},
n
P
a ∈ R+ , b, c, d, xk ∈ R∗+ , k = 1, n, Xn =
xk , m, p, r, s ∈ [1, +∞), such that
k=1
cXnm > d max16k6n xm
k , then
n
X
(aXnr + bxrk )s
(anr + b)s mp−rs+1 rs−mp
>
n
Xn
.
p
(cXnm − dxm
(cnm − d)p
k )
k=1
REFERENCES
[1] D.M. Bǎtineţu-Giurgiu, N. Stanciu, Inegalitǎţi geometrice n poligoane convexe, de tip Bergström-Mitrinović, Recreaţii Matematice, 2 (2011), 112–115.
[2] D.M. Bǎtineţu-Giurgiu, Aplicaţii la inegalitatea lui J. Radon, Gazeta Matematicǎ – seria B,
7-8-9 (2010), 359–362.
[3] D.M. Bǎtineţu-Giurgiu, N. Stanciu, Problem 11634, The American Mathematical Monthly,
119 (March 2012), 248.
[4] D.M. Bǎtineţu-Giurgiu, N. Stanciu, Nuevas generalizaciones y aplicaciones de la desigualdad
de Nesbitt, Revista Escolar de la Olimpiada Iberoamericana de Matematica 47 (nov. 2012–
feb. 2013).
“Matei Basarab” National College, Bucharest, Romania
“George Emil Palade” Secondary School, Buzǎu, Romania
E-mail: [email protected]