Question Number 196629 by universe last updated on 28/Aug/23 | ||
$${if}\:{xyz}=\mathrm{1},\:{prove} \\ $$$$\left(\frac{{x}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{{y}}{{y}−\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{{z}}{{z}−\mathrm{1}}\right)^{\mathrm{2}} \geqslant\mathrm{1}. \\ $$ | ||
Commented by universe last updated on 28/Aug/23 | ||
$$ \\ $$$${solution}\:{to}\:{question}\:#\mathrm{196519} \\ $$ | ||
Answered by universe last updated on 28/Aug/23 | ||
$$\:\:\:\:\:\:\:{let}\:\:\:{a}=\:\frac{{x}}{\mathrm{1}−{x}}\:\:,\:{b}=\:\frac{{y}}{\mathrm{1}−{y}}\:\:,\:\:{c}\:\:=\:\:\frac{{z}}{\mathrm{1}−{z}} \\ $$$$\:\:\:{abc}\:\:=\:\frac{{xyz}}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)\left(\mathrm{1}−{z}\right)}\:=\:\frac{\mathrm{1}}{\left.\left(\mathrm{1}−{x}\right)\mathrm{1}−{y}\right)\left(\mathrm{1}−{z}\right)} \\ $$$$\:\:\:{a}+\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\:,\:{b}+\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{1}−{y}}\:\:,\:{c}+\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{1}−{z}} \\ $$$$\left({a}+\mathrm{1}\right)\left({b}+\mathrm{1}\right)\left({c}+\mathrm{1}\right)\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)\left(\mathrm{1}−{z}\right)} \\ $$$$\left({a}+\mathrm{1}\right)\left({b}+\mathrm{1}\right)\left({c}+\mathrm{1}\right)\:=\:{abc} \\ $$$${abc}\:=\:{abc}+{ab}+{bc}+{ca}+{a}+{b}+{c}+\mathrm{1} \\ $$$${ab}+{bc}+{ca}+{a}+{b}+{c}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{2}\left({ab}+{bc}+{ca}\right)+\mathrm{2}\left({a}+{b}+{c}\right)\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ca}\right)+\mathrm{2}\left({a}+{b}+{c}\right)+\mathrm{2}\:=\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} +\mathrm{2}\left({a}+{b}+{c}\right)+\mathrm{2}\:=\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\:\:\left({a}+{b}+{c}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\:=\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\:\:\:\:{now}\:\:\left({a}+{b}+{c}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\:\geqslant\mathrm{1} \\ $$$$\:\:\:\:{so}\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant\mathrm{1} \\ $$$$\:\:\:\:\left(\:\frac{{x}}{\mathrm{1}−{x}}\right)^{\mathrm{2}} +\left(\frac{{y}}{\mathrm{1}−{y}}\right)^{\mathrm{2}} +\left(\frac{{z}}{\mathrm{1}−{z}}\right)^{\mathrm{2}} \geqslant\:\mathrm{1}\: \\ $$ | ||
Commented by mr W last updated on 28/Aug/23 | ||
$${great}!\:{thanks}! \\ $$ | ||