Proof by cases

#proof by cases#parity#absolute value

Statement

Prove that for all real $x$ , $|x|\ge x$ .

Hint

Split into $x\ge0$ and $x<0$ .

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Case 1: $x\ge0$ . $|x|=x$ , so $|x|\ge x$ .

Case 2: $x<0$ . $|x|=-x>0>x$ , so $|x|>x$ , hence $|x|\ge x$ .

Conclusion. In all cases $|x|\ge x$ .