Advanced book on Mathematics Olympiad

30 2 Algebra

And finally a more challenging problem from the 64th W.L. Putnam Mathematics
Competition.

Example.Letfbe a continuous function on the unit square. Prove that

∫ 1

0

(∫ 1

0

f (x, y)dx

) 2

dy+

∫ 1

0

(∫ 1

0

f (x, y)dy

) 2

dx

≤

(∫ 1

0

∫ 1

0

f (x, y)dxdy

) 2

+

∫ 1

0

∫ 1

0

f (x, y)^2 dxdy.

Solution.To make this problem as simple as possible, we prove the inequality for a
Riemann sum, and then pass to the limit. Divide the unit square inton^2 equal squares, then
pick a point(xi,yj)in each such square and defineaij =f(xi,yj),i, j= 1 , 2 ,...,n.
Written for the Riemann sum, the inequality becomes

1

n^3

∑

i

⎛

⎜

⎝

⎛

⎝

∑

j

aij

⎞

⎠

2

+

⎛

⎝

∑

j

aji

⎞

⎠

2 ⎞

⎟

⎠≤

1

n^4

⎛

⎝

∑

ij

aij

⎞

⎠

2

+

1

n^2

⎛

⎝

∑

ij

aij^2

⎞

⎠.

Multiply this byn^4 , then move everything to one side. After cancellations, the inequality
becomes

(n^2 − 1 )^2

∑

ij

a^2 ij+

∑

i   =k,j    =l

aijakl−(n− 1 )

∑

ij k,j  =k

(aijaik+ajiaki)≥ 0.

Here we have a quadratic function in theaij’s that should always be nonnegative. In
general, such a quadratic function can be expressed as an algebraic sum of squares, and
it is nonnegative precisely when all squares appear with a positive sign. We are left with
the problem of representing our expression as a sum of squares. To boost your intuition,
look at the following tableau:

a 11 ··· ··· ··· ··· ···a 1 n

..

.

... ..

.

... ..

.

... ..

.

··· ···aij ···ail··· ···

..

.

... ..

.

... ..

.

... ..

.

··· ···akj···akl··· ···

..

.

... ..

.

... ..

.

... ..

.

an 1 ··· ··· ··· ··· ···ann

The expression

(aij+akl−ail−akj)^2