Question Number 196519 by mr W last updated on 26/Aug/23
$${if}\:{xyz}=\mathrm{1},\:{prove} \\ $$$$\left(\frac{{x}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{{y}}{{y}−\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{{z}}{{z}−\mathrm{1}}\right)^{\mathrm{2}} \geqslant\mathrm{1}. \\ $$
Answered by mahdipoor last updated on 27/Aug/23
$${f}\left({x},{y},{z}\right)=\left(\frac{{x}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{{y}}{{y}−\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{{z}}{{z}−\mathrm{1}}\right)^{\mathrm{2}} \\ $$$${g}\left({x},{y},{z}\right)={xyz}=\mathrm{1} \\ $$$${if}\:{p}=\left({X},{Y},{Z}\right)\:{is}\:{extermum}\:{f}\:{in}\:{D}_{{f}} ={D}_{{g}} \\ $$$$\Rightarrow\overset{\rightarrow} {\bigtriangledown}{f}\left({p}\right)=\lambda\overset{\rightarrow} {\bigtriangledown}{g}\left({p}\right) \\ $$$${I}\:\:\:\:\:\Rightarrow\mathrm{2}\left(\frac{{X}}{{X}−\mathrm{1}}\right)\left(\frac{−\mathrm{1}}{\left({X}−\mathrm{1}\right)^{\mathrm{2}} }\right)=\lambda\left({YZ}\right) \\ $$$${II}\:\:\:\Rightarrow\mathrm{2}\left(\frac{{Y}}{{Y}−\mathrm{1}}\right)\left(\frac{−\mathrm{1}}{\left({Y}−\mathrm{1}\right)^{\mathrm{2}} }\right)=\lambda\left({XZ}\right) \\ $$$${III}\Rightarrow\mathrm{2}\left(\frac{{Z}}{{Z}−\mathrm{1}}\right)\left(\frac{−\mathrm{1}}{\left({Z}−\mathrm{1}\right)^{\mathrm{2}} }\right)=\lambda\left({XY}\right) \\ $$$${IV}\:\Rightarrow{XYZ}=\mathrm{1} \\ $$$${X},{Y}=\mathrm{0}\:,\:{Z}=\infty\:\:{or}\:{etc} \\ $$$$\Rightarrow{f}\left(\mathrm{0},\mathrm{0},\infty\right)=\mathrm{1}={minimum} \\ $$$${f}\left(\left({x},{y},{z}\right)\in{D}_{{g}} \right)>\mathrm{1} \\ $$
Commented by AST last updated on 27/Aug/23
$${f}\left(\frac{\mathrm{1}}{\mathrm{4}},−\mathrm{2},−\mathrm{2}\right)=\mathrm{1} \\ $$