Algebra 375and the inequality is proved.
(64th W.L. Putnam Mathematical Competition, 2003)
124.The inequality from the statement is equivalent to
0 < 1 −(a+b+c)+ab+bc+ca−abc <1
27
,
that is,
0 <( 1 −a)( 1 −b)( 1 −c)≤1
27
.
From the triangle inequalitiesa+b>c,b+c>a,a+c>band the condition
a+b+c=2 it follows that 0< a,b,c <1. The inequality on the left is now evident,
and the one on the right follows from the AM–GM inequality
√ (^3) xyz≤x+y+z
3
applied tox= 1 −a,y= 1 −b,z= 1 −c.
125.It is natural to try to simplify the product, and for this we make use of the AM–GM
inequality:
∏^25
n= 1
(
1 −
n
365)
≤
[
1
25
∑^25
n= 1(
1 −
n
365)]^25
=
(
352
365
) 25
=
(
1 −
13
365
) 25
.
We now use Newton’s binomial formula to estimate this power. First, note that
(
25
k)(
13
365
)k
≥(
25
k+ 1)(
13
365
)k+ 1
,since this reduces to
13
365≤
k+ 1
25 −k,
and the latter is always true for 1≤k ≤ 24. For this reason if we ignore the part
of the binomial expansion beginning with the fourth term, we increase the value of the
expression. In other words,
(
1 −13
365
) 25
≤ 1 −
(
25
1
)
13
365
+
(
25
2
)
132
3652
= 1 −
65
73
+
169 · 12
632
<
1
2
.
We conclude that the second number is larger.
(Soviet Union University Student Mathematical Olympiad, 1975)