Proofs by Case (2)

Write clear and complete proofs by cases for the mathematical statement.

(2) If x is a real number, then |x - 1| + |x + 5| >= 6.

Proof. Assume that x is a real number. Consider the following cases:

Case 1: Assume that x  < -5. In this case x - 1 < 0, and x + 5 < 0. So |x -1| = -(x - 1) = 1 - x, and |x + 5| = -(x + 5) = -5 - x. Then

|x - 1| + |x + 5| = 1 - x - 5 - x = -4 - 2x > -4 -2(-5) = -4 + 10 = 6.

So in this case |x - 1| + |x + 5| >= 6.

Case 2: Assume that -5 =< x < 1. In this case x - 1 < 0, and x + 5 >= 0.

So |x - 1| = -(x - 1) = 1 - x, and |x + 5| = x + 5. Then

|x - 1| + |x + 5| = 1 - x + x + 5 = 6 >= 6.

Case 3: Assume that x >= 1. In this case x - 1 < 0, and x + 5 >= 0. So

|x - 1| = x -1 and |x + 5| = x + 5. Then

|x - 1| + |x+5| = x - 1 + x + 5 = 2x + 4 >= 2(1) + 4 = 6

So in all cases |x - 1| + |x + 5| >= 6.

QED