Page 36 - RMGO 6

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36 Dorin MARGHIDANU

III. 1) If x ≥ 0, y < 0, x + y ≥ 0, then the given inequalities become
2
x + y 2
2
2
2
2
2
x + y ≤ ≤ x − y ⇔ x − y ≤ x + y ≤ (x − y) ,
x − y
obviously true.
III. 2) If x ≥ 0, y < 0, x + y < 0, then the given inequalities become
2
x + y 2
2
2
2
2
2
−(x + y) ≤ ≤ x − y ⇔ −x + y ≤ x + y ≤ (x − y) ,
x − y
obviously true.
IV. 1) If x < 0, y ≥ 0, x + y ≥ 0, then the given inequalities become
2
2
x + y 2 x + y 2
x + y ≤ − ≤ −x + y ⇔ x + y ≤ ≤ y − x
x − y y − x
2
2
2
2
2
⇔ y − x ≤ x + y ≤ (y − x) ,
obviously true.
IV. 2) If x < 0, y ≥ 0, x + y < 0, then the given inequalities become
2
2
−x − y 2 x + y 2
−(x + y) ≤ ≤ −x + y ⇔ −(x + y) ≤ ≤ y − x
x − y y − x
2
2
2
2
2
⇔ x − y ≤ x + y ≤ (y − x) ,
obviously true.
Resuming carefully the above proof, we obtain the following reformulation of
the statement from the Lemma 1.
Corollary 1. a) If the real numbers x and y have the same sign and x 6= y, then
x |x| − y |y|
|x + y| = = |x| + |y| . (2)
x − y
b) If the real numbers x and y have different signs, then
x |x| − y |y|
|x + y| < < |x| + |y| . (3)
x − y
Remark 1. As an illustration of relations (2) and (3), for x = −5 and y = −3
x |x| − y |y| −5 |−5| + 3 |−3| −25 + 9
we have |x + y| = |−5 − 3| = 8, = = = 8,
x − y −5 + 3 −2
|x| + |y| = |−5| + |−3| = 8,
x |x| − y |y|
and for x = 5 and y = −3 we have |x + y| = |5 − 3| = 2, =
x − y
5 |5| + 3 |−3| 34
= = 4,25, |x| + |y| = |5| + |−3| = 8 s , i 2 < 4,25 < 8.
5 + 3 8