Page 16 - MATINF Nr. 7

16 D. M˘arghidanu

 ‹
1 1 1 A n (a 1 , a 2 , . . . , a n )
Proof. The inequality A n (a 1 , a 2 , . . . , a n ) · A n , , . . . , ≥ is even the
b 1 b 2 b n A n (b 1 , b 2 , . . . , b n )
left inequality in (1) - which (we have seen) occurs even without any specific ordering condition.
Because the sequences (a 1 , a 2 , . . . , a n ) and (b 1 , b 2 , . . . , b n ) are similarly ordered, it follows
‹

1 1 1
that the sequences (a 1 , a 2 , . . . , a n ) and , , . . . , are inversely ordered, so by using
 ‹
b 1 b 2 b n
1 1 1
Chebishsev’s Inequality for this situation, we obtain A n (a 1 , a 2 , . . . , a n ) · A n , , . . . , ≥
 ‹ b 1 b 2 b n
a 1 a 2 a n
A n , , . . . , .
b 1 b 2 b n
 ‹
A n (a 1 , a 2 , . . . , a n ) a 1 a 2 a n
Remark 3. In general, the numbers and A n , , . . . , cannot be
A n (b 1 , b 2 , . . . , b n ) b 1 b 2 b n
compared in the case of synchronous sequences. For example, for similarly ordered sequences
(a 1 , a 2 , a 3 ) = (1, 2, 3) and (b 1 , b 2 , b 3 ) = (2, 3, 4), we will have
 ‹  ‹
A 3 (a 1 , a 2 , a 3 ) 2 a 1 a 2 a 3 1 2 3
= = 0,666... > A 3 , , = A 3 , , = 0,638...,
A 3 (b 1 , b 2 , b 3 ) 3 b 1 b 2 b 3 2 3 4

while for similarly ordered sequences (a 1 , a 2 , a 3 ) = (1, 1, 2) and (b 1 , b 2 , b 3 ) = (1, 2, 2), we will
have
 ‹  ‹  ‹
A 3 (a 1 , a 2 , a 3 ) 4 a 1 a 2 a 3 1 1 2 1
= = 0,8 < A 3 , , = A 3 , , = A 3 1, , 1 = 0,833....
A 3 (b 1 , b 2 , b 3 ) 5 b 1 b 2 b 3 1 2 2 2

If we explicitly express the means in relation (1) and multiply everywhere by n, we obtain:

Corollary 1. If a i ≥ 0, b i > 0, ∀ i = 1, n are such that the sequences (a 1 , a 2 , . . . , a n ) and
(b 1 , b 2 , . . . , b n ) are asynchronous, then
 ‹
a 1 + a 2 + . . . + a n 1 1 1 1
≤ · (a 1 + a 2 + . . . + a n ) · + + . . . +
b 1 + b 2 + . . . + b n n 2 b 1 b 2 b n
 ‹
1 a 1 a 2 a n
≤ · + + . . . + . (8)
n b 1 b 2 b n

Remark 4. The inequality between the extremes in (1), written in the form

a 1 a 2 a n a 1 + a 2 + . . . + a n
+ + . . . + ≥ n · , (9)
b 1 b 2 b n b 1 + b 2 + . . . + b n

with (a 1 , a 2 , . . . , a n ) and (b 1 , b 2 , . . . , b n ) (a i ≥ 0, b i > 0, (∀)i = 1, n) asynchronous sequences,
suggests the comparison with two other known inequalities for sums of fractions, namely
Bergstr¨om’s Inequality and Radon’s Inequality (see [1], [2], [4], [5], [8]).

We recall here these fractional inequalities:

Proposition 6. a) (Bergstr¨om’s Inequality, 1952)
?
If n ∈ N , a i ≥ 0, b i > 0, ∀ i = 1, n, then

a 2 1 a 2 2 a 2 n (a 1 + a 2 + . . . + a n ) 2
+ + . . . + ≥ ,
b 1 b 2 b n b 1 + b 2 + . . . + b n