42
On an Inequality from the IMO 2008
Nikolai Nikolov and Svilena Hristova
The following problem is from the IMO 2008 :
2
2
2
x
y
z
(IMO 2008) Prove that (1 −
+
+
≥ 1 for
x)2
(1 − y)2
(1 − z)2
all real numbers x, y, z , eah dierent from 1, and satisfying xyz = 1.
Problem 2(a)
Replaing x, y, z respetively by x1 , y1 , z1 , the inequality beomes
1
1
1
+
+
≥ 1.
(1 − x)2
(1 − y)2
(1 − z)2
The aim of this note is to show that this inequality remains true for
three or more variables. More preisely, we have the following.
Let n ≥ 2 be an integer and let x1 , x2 , . . . , xn be real numbers,
n
1
eah dierent from 1, and satisfying x1 x2 · · · xn = 1. Let Sn =
.
(1 − x )2
P
Proposition 1
i=1
(a)
(b)
()
(d)
i
If n = 2, then Sn ≥ , with equality if and only if x = y = −1.
If n = 3, then Sn ≥ 1, with equality if and only if x + y + z = 3.
If n = 4, then Sn ≥ 1, with equality if and only if x = y = z = t = −1.
If n ≥ 5, then Sn > 1. The inequality is sharp.
2
2
Proof : Clearing the frations in (a), the inequality beomes x + y ≥ 2xy ,
2
that is, (x − y) ≥ 0.
Clearing the frations in (b), the inequality beomes (x+y+z−3)2 ≥ 0.
To prove () and (d), we shall use the following result (see [2℄, and also
the remark at the end of [1℄) :
If y1 , y2 , . . . , yn are positive real numbers, 1 − n ≤ α < 0,
n
n
and
yi = λn , then
(1 + yi )α ≥ min {1 , n(1 + λ)α }. The
i=1
i=1
inequality is sharp, with equality if and only if n(1 + λ)α ≤ 1 and
y1 = y2 = · · · = yn = λ.
This result implies () and (d) by setting α = −2, λ = 1, yi = |xi |, and
using the fat that (1 −1x )2 ≥ (1 +1y )2 .
i
i
A more diret approah for proving () and (d) is to use the inequality
1
1
1
+
≥
for a, b ≥ 0. This inequality holds sine it is
(1 + a)2
(1 + b)2
1 + ab
1
2
Q
Copyright
c 2010
P
Canadian Mathematial Soiety
Crux Mathematiorum with Mathematial Mayhem, Volume 36, Issue 1
43
equivalent to the obvious inequality (ab − 1)2 + ab(a − b)2
follows immediately :
X
4
i=1
≥ 0.
Then ()
1
1
1
≥
+
= 1.
(1 − xi )2
1 + |x1 x2 |
1 + |x3 x4 |
To prove (d), it is enough to observe that
X
i=1
X
€1 + 1|x |Š ≥
X 1
1
+
€1 + |x |Š
1 + |x x |
n
1
(1 − xi )2
n
≥
2
i=1
i
n
2
1
2
i=3
i
and then apply indution on n.
≥
€1 + |x1 x |Š
2
1
X
n
+
2
i=3
€1 + 1|x |Š
2
i
More is true when all the variables are positive.
With notation as in Proposition 1, if additionally x1 , . . . , xn
n
1
are positive, then
> 1. The inequality is sharp.
(1 − x )2
P
Proposition 2
i=1
€
i
Š
For n = 2, by learing frations, the inequality beomes x + y > 2,
2
√
that is, x − √y > 0. It remains only to note that x = y implies that
x = y = 1.
For n = 3 we know that strit inequality holds in Proposition 1(b) when
(x + y + z − 3)2 > 0. In the present ase it then suÆes to observe that
√
x + y + z ≥ 3 xyz = 3, with equality if and only if x = y = z = 1.
If n ≥ 4, the inequality follows from Proposition 1, parts () and (d).
Finally, to see that the inequality is sharp, set x1 = · · · = xn−1 = j ,
1
xn = n−1 , and let j → ∞.
j
Proof :
3
Referenes
[1℄ O. Mushkarov and N. Nikolov, Some generalizations of an inequality
from IMO 2001, CRUX Mathematiorum with Mathematial Mayhem,
Vol. 28, No. 5 (2002) pp. 308-312.
[2℄ O. Mushkarov and N. Nikolov, Variations on an inequality from IMO
2001, Mathematis and Eduation in Mathematis, Vol. 32 (2003)
pp. 323-327. (arXiv:math.HO/0605380v1)
Nikolai Nikolov
[email protected]
Institute of Mathematis and Informatis
Bulgarian Aademy of Sienes
1113 Soa, Bulgaria
Svilena Hristova
[email protected]