Proof of Triangle Inequality (Absolute Values)

Solution: Triangle Inequality

Statement: For all real numbers x and y,

|x + y| ≤ |x| + |y|

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(a) Verification

Case (i): x = 2, y = 3

|x + y| = |2 + 3| = |5| = 5
|x| + |y| = |2| + |3| = 2 + 3 = 5
∴ |x + y| = |x| + |y|

Case (ii): x = −2, y = −3

|x + y| = |−2 − 3| = |−5| = 5
|x| + |y| = |−2| + |−3| = 2 + 3 = 5
∴ |x + y| = |x| + |y|

Case (iii): x = −2, y = 3

|x + y| = |−2 + 3| = |1| = 1
|x| + |y| = |−2| + |3| = 2 + 3 = 5
∴ |x + y| ≤ |x| + |y|

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(b) Proof for all real numbers

We prove the inequality by considering different cases.

Case 1: x ≥ 0, y ≥ 0

|x + y| = x + y = |x| + |y|
Hence, |x + y| ≤ |x| + |y|.

Case 2: x ≤ 0, y ≤ 0

|x + y| = −(x + y) = −x − y = |x| + |y|
Hence, |x + y| ≤ |x| + |y|.

Case 3: x and y have opposite signs

In this case, the sum x + y is reduced in magnitude. Therefore,

|x + y| < |x| + |y|

Hence, the inequality holds.

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Therefore, the Triangle Inequality holds for all real numbers x and y.