THREE PROBLEMS IN ONE
Finbarr Holland
1. Introdution
Sometime over the summer I reeived a postard from Gordon
Lessells (Deputy Leader), Kevin Huthinson (Team Leader) and
Pat MCarthy (Observer), who were with the Irish Team at the
1999 IMO in Buharest, bearing the following rypti message:
Find the smallest onstant C suh that
n !4
X
2
2
:
xi
xi xj (xi + xj ) C
i=1
1i<jn
X
Can you nd the two{line Chinese proof or the Irish solution?
When I got the ard I realized that the problem posed was one of
the six problems seleted for the IMO, and was triply intrigued
by it. Firstly, it didn't appear to fall easily by any of the standard
methods that IMO partiipants would know, and therefore struk
me as possessing an element of novelty and, as J. E. Littlewood has
said, \the disovery of a new inequality is a ause for elebration."
Seondly, beause I had no idea what would pass for an aeptable
\Chinese Solution." Thirdly, I wondered how I ould possibly be
expeted to redisover the \Irish Solution."
I suspeted that Pat MCarthy{a former student of mine{
might have had a hand in oming up with the \Irish Solution."
But ould I redisover it? Did the link between Pat and myself
provide a lue? Was the \Irish Solution" somehow \geneti"?
When I met him at the reent IMS meeting in TCD, Gordon
eliited my response to these questions. I had to onfess that I
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74
IMS Bulletin 43, 1999
hadn't found any solution, but that I would like to think about
it some more. During the meeting I found the seond solution
outlined below, and when I ommuniated it to himself and Kevin
Huthinson, who was also at the IMS meeting, Kevin informed
that it was essentially the \Irish Solution," disovered by Pat and
himself, who were unhappy with the inelegane of the proposed
solutions disussed at the Jury Meetings, and were motivated to
produe a nier one.
Gordon later told me that a solution similar to the \Irish
Solution" was also found by an Iranian student during the ontest,
and desribed the \Chinese solution" outlined below, whih is
based on an idea that a Chinese student used during the ontest,
and was elaborated later on by one of the oordinators.
Here's the formulation of the problem that the students were
given on the rst morning of the Olympiad:
Problem 1 Let n 2 be a xed integer.
1. Find the least onstant C suh that
X
1i<jn
i j (x2i + x2j ) C
x x
n
X
i=1
!4
x
i
;
for all non-negative real numbers x1 , x2 ,: : : , xn .
2. Determine when equality ours for this value of C .
2. The \Chinese Solution"
Let
M
=
n
X
2
i
i=1
x :
Then it is lear that
X
1i<jn
i j (x2i + x2j ) M
x x
X
1i<jn
i j
x x ;
with equality if, and only if, at worst all but two of the variables
are zero.
Three Problems in One
75
Now, as is well known, 4ab (a + b)2 if a and b are real,
with equality P
if, and only if, a = b. Hene, applying this with
a = M, b = 2
1i<jn xi xj we see that
M
0
X
1i<jn
i j
x x
12
n
X
1 X 2
xi + 2
xi xj A
8 i=1
1i<jn
n
1 X 2
=
( x)
8 i=1 i
1
=
8
n
X
i=1
!2
!4
i
x
;
with equality if, and only if,
n
X
2=2 X
i
i=1
x
n
1 X
i j = 2 ( xi )2 :
i=1
1i<jn
x x
That is, if, and only if, two of the variables are equal and the rest
are zero.
Thus, C = 1=8, and equality holds if, and only if, all of the
variables are zero or at most two are non-zero and equal.
3. The \Irish Solution"
This uses the following easily established identities:
8ab(a2 + b2 ) = (a + b)4
and
(a
n
n X
X
3
2
2
xi
xi
xi xj (xi + xj ) =
i=1 i=1
1i<jn
X
)4 ;
(1)
b
n
X
4
i
i=1
x :
(2)
The rst of these tells us that C 1=8 and that equality holds in
ase n = 2 if, and only if, the variables are equal.
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IMS Bulletin 43, 1999
The seond identity suggests an alternative formulation of
the problem, viz., nd the smallest positive onstant C suh that
n
X
3 (1
i
i=1
i) C
x
x
subjet to the onstraints that 0 xi , i = 1, 2, : : : , n, and
n
X
i=1
then
x
i = 1:
As a rst shot at solving this we observe that if 0 x 1,
2 (1
x
x
) = 4( )( )(1 x)
2 2
x + x + 1 x 3
4 2 2 3
4
= ;
27
by the Arithmeti{Geometri Mean inequality, with equality holding if, and only if, x = 2=3. Hene,
x
n
X
3 (1
i
i=1
x
x
2
x (1
i ) 1max
in i
x
274
i)
x
n
X
i=1
i
x
:
This establishes that C 4=27, but it is lear that the bound is
rude and not attainable.
To rene this argument, we distinguish three ases: Case
(A), in whih we suppose all of the xi belong to the interval
[0; 1=2); Case (B), in whih we suppose that exatly one of the
xi belongs to (1=2; 1℄, and Case (C), in whih we suppose that
preisely two of the xi are equal to 1=2. We proeed in the knowledge that these ases exhaust all possibilities.
Three Problems in One
77
Sine x(1 x) y (1 y ) = (x y )(1 x y ) it is easy
to see that x(1 x) is stritly inreasing on [0; 1=2℄. Hene, too,
2
x (1
x) is stritly inreasing on [0; 1=2℄, so that if 0 x < 1=2,
then x2 (1 x) < 1=8, and so, in Case (A),
n
X
3 (1
i
i=1
x
2
x (1
i ) 1max
in i
x
= max x2i (1
1in
<
i)
x
i)
n
X
i=1
i
x
x
1
:
8
Suppose next that one of the xi belongs to (1=2; 1℄. Denote
this by a. The sum of the remaining n 1 variables (whih we
now relabel as y1 , y2 , : : :, yn 1 ) is 1 a < 1=2, so that eah
yi 2 [0; 1
a℄ [0; 1=2). It follows that
y
2
i (1
i ) (1
y
a
)2 a;
i
= 1; 2; : : : ; n
1:
Hene
n
X
3 (1
i
i=1
x
i ) = a3 (1
x
)+
nX1
i=1
3
i (1
i)
y
y
2 (1
1in 1 i
a3 (1 a) + (1 a)3 a
1
=
(2a 1)4
<
a
3 (1
a
a
) + max
y
i)
y
nX1
i=1
i
y
8
1
;
8
by identity (1).
This disposes of Case (B); we are left with Case (C), whih
is trivial.
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IMS Bulletin 43, 1999
The
P upshot of this analysis is that, if 0
, and ni=1 xi = 1, then
i , i = 1, 2,: : : ,
x
n
n
X
3 (1
i
i=1
x
1
i) 8 ;
x
and that the inequality is strit unless two of the variables are
equal to 1=2 and the rest are zero.
4. A Problem for the Reader
Let n 2 be a positive integer and p a positive number. Determine (a) the best onstant Cp suh that
n !p+2
X
p
p
;
xi
xi xj (xi + xj ) Cp
i=1
1i<jn
X
for all nonnegative real numbers x1 , x2 ,: : : , xn , and (b) the ases
of equality.
Finbarr Holland
Mathematis Department
University College
Cork
Ireland.