Page 37 - RMGO 6

A refinement of the Modulus inequality 37

Next we derive an extension of inequality (1), even with its help, for the case
of three variables.

Proposition 1 (A refinement of Modulus inequality in three variables,
[1], b)). For any x, y, z ∈ R with x 6= y 6= z 6= x, the following inequalities hold
1 Å x |x| − y |y| y |y| − z |z| z |z| − x |x| ã
|x + y + z| ≤ · + + ≤ |x|+|y|+|z| . (4)
2 x − y y − z z − x
Proof. Using the Modulus inequality and the left inequality from (1), we have

1
|x + y + z| = · |(x + y) + (y + z) + (z + x)|
2
1
≤ · (|x + y| + |y + z| + |z + x|)
2
Å ã
1 x |x| − y |y| y |y| − z |z| z |z| − x |x|
≤ · + + .
2 x − y y − z z − x
On the other hand, using the right inequality from (1), we obtain

1 Å x |x| − y |y| y |y| − z |z| z |z| − x |x| ã
· + +
2 x − y y − z z − x
1
≤ · [(|x| + |y|) + (|y| + |z|) + (|z| + |x|)]
2
= |x| + |y| + |z| .

Similarly, we obtain the following general refinement.

Proposition 2 (A refinement of Modulus inequality - the general case,
[2]). For any x 1 , x 2 , . . . , x n ∈ R with x 1 6= x 2 6= . . . 6= x n 6= x 1 , the following
inequalities hold

n n n

X 1 X x k |x k | − x k+1 |x k+1 | X
x k ≤ · ≤ |x k | , (5)

2

x k − x k+1
k=1 k=1 k=1
where x n+1 = x 1 .
Proof. After an easy preparation, using n times the Modulus inequality and then
the left inequality from Lemma 1, we have

n
X 1
x k = · |(x 1 + x 2 ) + (x 2 + x 3 ) + . . . + (x n + x n+1 )|

2

k=1
1
≤ · (|x 1 + x 2 | + |x 2 + x 3 | + . . . + |x n + x n+1 |)
2
n
1 X x k |x k | − x k+1 |x k+1 |
≤ · .
2 x k − x k+1
k=1