Page 16 - MATINF Nr. 6

16 D. M˘arghidanu

relations which, - by multiplying part by part, lead to:

(G−x 1 )+(G−x 2 )+...+(G−x ) G k kG−S k G k
k
e G ≤ ⇔ e G ≤ . (2)
x 1 · x 2 · . . . · x n P k
From (1) and the right inequality in Lemma 2, we obtain:
x −G
x k+1 k+1
G ≤ x k+1 → ≤ e G (3 1 )
G
x k+2 x k+2 −G
G ≤ x k+2 → ≤ e G (3 2 )
G
. . .

x n x n −G
G ≤ x n → ≤ e G (3 n−k )
G
relations which - by multiplying part by part, turn to:

(x +...+ x n )−(n−k)·G
x k+1 · x k+2 · ... · x n k+1
≤ e G . (3)
G n−k
x k+1 · x k+2 · . . . · x n x k+1 · x k+2 · . . . · x n
But, = √ =
G n−k n n−k
x 1 · x 2 · . . . · x n
k
(x 1 · x 2 · . . . · x n ) · (x k+1 · x k+2 · . . . · x n )
n
x k+1 · x k+2 · . . . · x n
= k = =
(x 1 · x 2 · . . . · x n ) 1− n x 1 · x 2 · . . . · x n
k
(x 1 · x 2 · . . . · x n ) n G k
= = , (4)
x 1 · x 2 · . . . · x k P k
So (3) can be restated as:
G k (x k+1 +...+x n )−(n−k)·G
≤e G . (5)
P k
From (2) and (5) by transitivity, it results:

k · G − S k ≤ (x k+1 + . . . + x n ) − (n − k) · G ⇔ k · G + (n − k) · G ≤ S k + (x k+1 + . . . + x n )

⇔ n · G ≤ S n ⇔ G ≤ A.

The equality is obtained when we have equality everywhere in (1) so when x 1 = x 2 = . . . =
x n .

References

[1] D. M˘arghidanu O demonstrat¸ie inductiv˘a a Inegalit˘at ,ii mediilor, RMGO, Anul IV , nr. 1,
2020, pp. 33-35, on-line, http://rmgo.upit.ro/RMGO4/index.html#p=32.