Question Number 157168 by mathocean1 last updated on 20/Oct/21
$${x}\:,\:{y}\:{and}\:{z}\:{are}\:{numbers}. \\ $$$${Show}\:{that}\:{max}\left({x},\:{y}\right)=\frac{{x}+{y}+\mid{x}−{y}\mid}{\mathrm{2}}\:{and}\:{min}\left({x},{y}\right)=\frac{{x}+{y}−\mid{x}−{y}\mid}{\mathrm{2}} \\ $$$${then}\:{find}\:{a}\:{formula}\:{for}\: \\ $$$${max}\left({x},{y},{z}\right). \\ $$
Answered by puissant last updated on 20/Oct/21
$$\rightarrow{max}\left({x};{y}\right)=\frac{\mathrm{1}}{\mathrm{2}}{max}\left(\mathrm{2}{x};\mathrm{2}{y}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{max}\left({x}−{y}+{x}+{y}\:;\:{y}−{x}+{x}+{y}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{y}+{max}\left({x}−{y}\:;\:{y}−{x}\right)\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{y}+\mid{x}−{y}\mid\right)\:=\:\frac{{x}+{y}+\mid{x}−{y}\mid}{\mathrm{2}} \\ $$$$\rightarrow{min}\left({x};{y}\right)=\frac{\mathrm{1}}{\mathrm{2}}{min}\left(\mathrm{2}{x};\mathrm{2}{y}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{min}\left({x}−{y}+{x}+{y}\:;\:{y}−{x}+{x}+{y}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{y}+{min}\left({x}−{y}\:;\:{y}−{x}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{y}−\mid{x}−{y}\mid\right)=\frac{{x}+{y}−\mid{x}−{y}\mid}{\mathrm{2}} \\ $$$$\rightarrow\:{max}\left({x}\:;\:{y}\:;\:{z}\right)=\:{max}\left({max}\left({x};{y}\right);{z}\right) \\ $$$$ \\ $$$$=\frac{{max}\left({x};{y}\right)+{z}+\mid{max}\left({x};{y}\right)−{z}\mid}{\mathrm{2}} \\ $$$$=\frac{\frac{{x}+{y}+\mid{x}−{y}\mid}{\mathrm{2}}+{z}+\mid\frac{{x}+{y}+\mid{x}−{y}\mid}{\mathrm{2}}−{z}\mid}{\mathrm{2}}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:…………\mathscr{L}{e}\:{puissant}……….. \\ $$
Commented by mathocean1 last updated on 22/Oct/21
$${thanks}\:{le}\:{puissant}. \\ $$
Commented by puissant last updated on 24/Oct/21
$${de}\:{rien}… \\ $$