Real Analysis 5492 m(A)IA+ 2 m(B)IB≥IA+IB.Sincem(A)+m(B)=1, this inequality translates into
(
m(A)−
1
2
)
(IA−IB)≥ 0 ,
which would be true ifIA≥IB. However, if this last assumption does not hold, we can
return to the term that we neglected, and use the triangle inequality to obtain
∫
A∫
B|f(x)+f(y)|dxdy≥∫
A∫
B|f(x)|−|f(y)|dxdy=m(A)IB−m(B)IA.The inequality from the statement would then follow from
2 m(A)IA+ 2 m(B)IB+ 2 m(A)IB− 2 m(B)IA≥IA+IB,which is equivalent to
(
m(A)−
1
2
)
(IA+IB)+m(B)(IB−IA)≥ 0.This is true since both terms are positive.
(64th W.L. Putnam Mathematical Competition, 2003)
486.Combining the Taylor series expansions
cosx= 1 −x^2
2!+
x^4
4!−
x^6
6!+
x^8
8!+···,
coshx= 1 +
x^2
2!+
x^4
4!+
x^6
6!+
x^8
8!+···,
we see that the given series is the Taylor series of^12 (cosx+coshx).
(TheMathematics GazetteCompetition, Bucharest, 1935)
487.Denote bypthe numerator and byqthe denominator of this fraction. Recall the
Taylor series expansion of the sine function,
sinx=x
1!−
x^3
3!+
x^5
5!−
x^7
7!+
x^9
9!+···.
We recognize the denominators of these fractions inside the expression that we are com-
puting, and now it is not hard to see thatpπ−qπ^3 =sinπ=0. Hencepπ=qπ^3 , and
the value of the expression from the statement isπ^2.
(Soviet Union University Student Mathematical Olympiad, 1975)