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For real numbers x and y, show that $$\max\{x,y\} = \frac{x+y +|x-y|}{2}$$

My attempt:

If $x\geq y$, then $$\frac{x+y+|x-y|}{2} = \frac{x+y+x-y}{2} = \frac{2x}{2} = x = \max\{x,y\}.$$

Similarly, if $x < y$, then $$\frac{x+y+|x-y|}{2} = \frac{x+y+y-x}{2} = \frac{2y}{2} = y = \max\{x,y\}.$$

Where do I go on from here?

Michael Rozenberg
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Adagio
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    You've done it. There were two cases and in both cases you showed $\max {x,y} = \frac{x+y+|x-y|}{2}$ – eepperly16 Feb 09 '17 at 04:58
  • Proof is correct. You could merge the two cases in one if you note that the proposition is symmetric in $x,y$. – dxiv Feb 09 '17 at 05:01
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    I like this proof. $x + y = \max(x,y) + \min(x,y)$ and $|x-y| = \max(x,y) - \min(x,y)$ add and you get $x+y+|x-y| = 2\max(x,y)$... – Steven Alexis Gregory Feb 09 '17 at 05:07

1 Answers1

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Yes, as mentioned in the comments, your proof is done. You need not show anything more. That is all you need to do, and your proof is correct and your reasoning sound. (Answer added as this is not an unanswered question).

S.C.B.
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