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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?

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    You've done it. There were two cases and in both cases you showed $\max \{x,y\} = \frac{x+y+|x-y|}{2}$ – 2017-02-09
  • 0
    Proof is correct. You could merge the two cases in one if you note that the proposition is symmetric in $x,y$. – 2017-02-09
  • 1
    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 $x+y+|x-y| = 2\max(x,y)$... – 2017-02-09

1 Answers 1

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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).