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Question Number 157168 by mathocean1 last updated on 20/Oct/21

$$x,yandzarenumbers.$$$$Showthatmax(x,y)=\frac{x+y+\mid x-y\mid }{2}andmin(x,y)=\frac{x+y-\mid x-y\mid }{2}$$$$thenfindaformulafor$$$$max(x,y,z).$$

Answered by puissant last updated on 20/Oct/21

$$\to max(x;y)=\frac{1}{2}max(2x;2y)$$$$=\frac{1}{2}max(x-y+x+y;y-x+x+y)$$$$=\frac{1}{2}(x+y+max(x-y;y-x)$$$$=\frac{1}{2}(x+y+\mid x-y\mid )=\frac{x+y+\mid x-y\mid }{2}$$$$\to min(x;y)=\frac{1}{2}min(2x;2y)$$$$=\frac{1}{2}min(x-y+x+y;y-x+x+y)$$$$=\frac{1}{2}(x+y+min(x-y;y-x))$$$$=\frac{1}{2}(x+y-\mid x-y\mid )=\frac{x+y-\mid x-y\mid }{2}$$$$\to max(x;y;z)=max(max(x;y);z)$$$$$$$$=\frac{max(x;y)+z+\mid max(x;y)-z\mid }{2}$$$$=\frac{\frac{x+y+\mid x-y\mid }{2}+z+\mid \frac{x+y+\mid x-y\mid }{2}-z\mid }{2}..$$$$$$$$............\mathcal{L}epuissant...........$$

Commented by mathocean1 last updated on 22/Oct/21

$$thankslepuissant.$$

Commented by puissant last updated on 24/Oct/21

$$derien...$$

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