Advanced book on Mathematics Olympiad

Algebra 377

and each of these is just the AM–GM inequality.
(Greek Team Selection Test for the Junior Balkan Mathematical Olympiad, 2005)

128.Denote the positive number 1−(a 1 +a 2 +···+an)byan+ 1. The inequality from
the statement becomes the more symmetric

a 1 a 2 ···anan+ 1

( 1 −a 1 )( 1 −a 2 )···( 1 −an)( 1 −an+ 1 )

≤

1

nn+^1

.

But from the AM–GM inequality,

1 −a 1 =a 2 +a 3 +···+an+ 1 ≥nn

√

a 2 a 3 ···an+ 1 ,

1 −a 2 =a 1 +a 3 +···+an+ 1 ≥nn

√

a 1 a 3 ···an+ 1 ,

···

1 −an+ 1 =a 1 +a 2 +···+an≥nn

√

a 1 a 2 ···an.

Multiplying thesen+1 inequalities yields

( 1 −a 1 )( 1 −a 2 )···( 1 −an+ 1 )≥nn+^1 a 1 a 2 ···an,

and the conclusion follows.
(short list of the 43rd International Mathematical Olympiad, 2002)

129.Trick number 1: Use the fact that

1 =

n− 1 +xj

n− 1 +xj

=(n− 1 )

1

n− 1 +xj

+

xj

n− 1 +xj

,j= 1 , 2 ,...,n,

to transform the inequality into

x 1

n− 1 +x 1

+

x 2

n− 1 +x 2

+···+

xn

n− 1 +xn

≥ 1.

Trick number 2: Break this into theninequalities

xj

n− 1 +xj

≥

x

1 −^1 n

j

x

1 −^1 n

1 +x

1 −^1 n

2 +···+x

1 −^1 n

n

,j= 1 , 2 ,...,n.

We are left withnsomewhat simpler inequalities, which can be rewritten as

x

1 −^1 n

1 +x

1 −^1 n

2 +x

1 −^1 n

j− 1 +x

1 −^1 n

j+ 1 +···+x

1 −^1 n

n ≥(n−^1 )x

−^1 n

j.

Trick number 3: Use the AM–GM inequality