Advanced book on Mathematics Olympiad

Real Analysis 527

and the inequality is proved.

(T. Popoviciu, solution published by Gh. Eckstein inTimi ̧soara Mathematics Gazette)

432.The fact that all sequences(anbn)nare convex implies that for any real numbera,

an+^1 bn+ 1 − 2 anbn+an−^1 bn− 1 ≥0. Hencebn+ 1 a^2 − 2 bna+bn− 1 ≥0 for alla. Viewing

the left-hand side as a quadratic function ina, its discriminant must be less than or equal

to zero. This is equivalent tob^2 n≤bn+ 1 bn− 1 for alln. Taking the logarithm, we obtain

that 2 lnbn≤lnbn+ 1 +lnbn− 1 , proving that the sequence(lnbn)nis convex.

433.We will show that the largest such constant isC=^12. For example, if we consider

the sequencea 1 =,a 2 =1,a 3 =, witha small positive number, then the condition

from the statement implies

C≤

1

2

·

( 1 + 2 )^2

1 + 2 ^2

.

Here if we let→0, we obtainC≤^12.

Let us now show thatC=^12 satisfies the inequality for all concave sequences. For

everyi, concavity forces the elementsa 1 ,a 2 ,...,aito be greater than or equal to the

corresponding terms in the arithmetic progression whose first term isa 1 and whoseith

term isai. Consequently,

a 1 +a 2 +···+ai≥i

(

a 1 +ai

2

)

.

The same argument repeated forai,ai+ 1 ,...,anshows that

ai+ai+ 1 +···+an≥(n−i+ 1 )

(

ai+an

2

)

.

Adding the two inequalities, we obtain

a 1 +a 2 +···+an≥i

(

a 1 +ai

2

)

+(n−i+ 1 )

(

ai+an

2

)

−ai

=i

a 1

2

+(n−i+ 1 )

an

2

+

(n− 1 )ai

2

≥

(

n− 1

2

)

ai.

Multiplying byaiand summing the corresponding inequalities for alligives

(a 1 +a 2 +···+an)^2 ≥

n− 1

2

(a 12 +a 22 +···+a^2 n).

This shows that indeedC=^12 is the answer to our problem.

(Mathematical Olympiad Summer Program, 1994)