OCTOGON MATHEMATICAL MAGAZINE
Vol. 17, No.1, April 2009, pp 291-293
ISSN 1222-5657, ISBN 978-973-88255-5-0,
www.hetfalu.ro/octogon
291
About one algebraic inequality
Šefket Arslanagić33
ABSTRACT. In this paper we present the error by the proof of one algebraic
inequality.
INTRODUCTION
In [2], p. 39, problem AQ.10.; in [3], p.15, problem 80. and in [4], p.5,
problem A1991-6. we have the next problem:
Prove the inequality
x2 y y 2 z z 2 x
+
+
≥ x2 + y 2 + z 2
z
x
y
(1)
for all positive numbers x, y, z.
Remark. In [2] and [3] in the place x, y, z we have a, b, c.
MAIN RESULTS
In [2], p. 69 and [3], p. 38, we have only this phrase as the solution:
With the Cauchy-Buniakowsky-Schwarzs inequality, we get:
2
2
2
x y y2z z2x
x z y2x z2y
+
+
+
+
≥ x2 + y 2 + z 2
z
x
y
y
z
x
(2)
But , I see not in what way to receive the proof of the given inequality (1)
from the inequality (2)?!
In [4], p. 41 we have this solution:
Without lost of generality we may assume that x ≥ y ≥ z > 0. We have
33
Received: 26.02.2009
2000 Mathematics Subject Classification. 26D15
Key words and phrases. Algebraic inequality; symmetric cyclic and homogeneous
inequality, contraexample.
292
Šefket Arslanagić
x2 y y 2 z z 2 x
+
+
≥ x2 + y 2 + z 2 ⇔ x3 y 2 + y 3 z 2 + z 3 x2 ≥ x3 yz + y 3 zx + z 3 xy
z
x
y
⇔ x3 y (y − z) + y 2 z 2 (y − z) + z 3 y 2 − 2yx + x2 − xyz y 2 − z 2 ≥ 0
⇔ (y − z) (x − z) x2 y + yz (x − y) + z 3 (x − y)2 ≥ 0
Then the last inequality holds.
Unfortunately, this proof is not complete. Why? The inequality (1) is cyclic
and homogeneous, but this inequality is not symmetric! We can not take
only that is x ≥ y ≥ z > 0.
It is not heawily give one contraexample, i.e. show that this inequality is not
exact, for example so x = 16, y = 1, z = 2 (x ≥ z ≥ y > 0); now we have:
256 1
1
+ + 64 ≥ 256 + 1 + 4 ⇔ 192 ≥ 261(?!)
2
8
8
4
In general, for x = n , y = 1, z = n; (n ∈ i; n ≥ 2) we have of (1):
n7 +
1
+ n6 ≥ n8 + 1 + n2 ⇔ n10 + n9 + 1 ≥ n11 + n5 + n3 ⇔
n3
⇔ n10 (n − 1) + n5 1 − n4 + n3 − 1 ≤ 0 ⇔
⇔ n10 (n − 1) − n5 n4 − 1 + (n − 1) n2 + n + 1 ≤ 0 ⇔
⇔ (n − 1) n10 − n5 (n + 1) n2 + 1 + n2 + n + 1 ≤ 0 ⇔
⇔ (n − 1) n10 + n2 + n + 1 − n8 − n7 − n6 − n5 ≤ 0 ⇔
8
7
6
⇔ (n − 1) n + n + n + n
5
n10 + n2 + n + 1
−1 ≤0⇔
n8 + n7 + n6 + n5
n6 + n2 + n + 1
≤0
⇔ n5 (n − 1) n3 + n2 + n + 1 n (n − 1) − 1 + 8
n + n7 + n6 + n5
what is not exact because n ≥ 2.
About one algebraic inequality
293
The inequality (1) not holds too for 0 < x ≤ y ≤ z, because for x = 1, y = 2,
z = 16, we get of (1):
1
1
+ 64 + 128 ≥ 1 + 4 + 256 ⇔ 192 ≥ 261,
8
8
what is not true.
Therefore, the inequality (1) not holds for all x, y, z > 0. This inequality
holds for x ≥ y ≥ z > 0.
REFERENCES
[1] Arslanagić, Š., Matematika za nadarene, Bosanska rijec, Sarajevo, 2005.
[2] Maftei, I.V., Popescu, P.G., Piticari, M., Lupu, C., Tataram, M.A.,
Inegalitati alese in matematica Inegalitati clasice, Editura Niculescu,
Bucuresti, 2005.
[3] Mortici, C., 600 de probleme, Editura Gil, Zalau, 2001.
[4] The Vietnamese Mathematical Olympiad (1990-2006), Selected Problems,
Education Publishing House, Hanoi-Vietnam, 2007.
University of Sarajevo
Faculty of Natural Sciences and Mathematics
Department of Mathematics
Zmaja od Bosne 35, 71000 Sarajevo, Bosnia and Herzegovina
E-mail: [email protected]