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Question Number 196629 by universe last updated on 28/Aug/23

if xyz=1, prove ((x/(x−1)))^2 +((y/(y−1)))^2 +((z/(z−1)))^2 ≥1.

$${if}\:{xyz}=\mathrm{1},\:{prove} \\ $$$$\left(\frac{{x}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{{y}}{{y}−\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{{z}}{{z}−\mathrm{1}}\right)^{\mathrm{2}} \geqslant\mathrm{1}. \\ $$

Commented by universe last updated on 28/Aug/23

solution to question #196519

$$ \\ $$$${solution}\:{to}\:{question}\:#\mathrm{196519} \\ $$

Answered by universe last updated on 28/Aug/23

let a= (x/(1−x)) , b= (y/(1−y)) , c = (z/(1−z)) abc = ((xyz)/((1−x)(1−y)(1−z))) = (1/((1−x)1−y)(1−z))) a+1 = (1/(1−x)) , b+1 = (1/(1−y)) , c+1 = (1/(1−z)) (a+1)(b+1)(c+1) = (1/((1−x)(1−y)(1−z))) (a+1)(b+1)(c+1) = abc abc = abc+ab+bc+ca+a+b+c+1 ab+bc+ca+a+b+c+1 = 0 2(ab+bc+ca)+2(a+b+c) + 2 = 0 a^2 +b^2 +c^2 +2(ab+bc+ca)+2(a+b+c)+2 = a^2 +b^2 +c^2 (a+b+c)^2 +2(a+b+c)+2 = a^2 +b^2 +c^2 (a+b+c+1)^2 +1 = a^2 +b^2 +c^2 now (a+b+c+1)^2 +1 ≥1 so a^2 +b^2 +c^2 ≥1 ( (x/(1−x)))^2 +((y/(1−y)))^2 +((z/(1−z)))^2 ≥ 1

$$\:\:\:\:\:\:\:{let}\:\:\:{a}=\:\frac{{x}}{\mathrm{1}−{x}}\:\:,\:{b}=\:\frac{{y}}{\mathrm{1}−{y}}\:\:,\:\:{c}\:\:=\:\:\frac{{z}}{\mathrm{1}−{z}} \\ $$$$\:\:\:{abc}\:\:=\:\frac{{xyz}}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)\left(\mathrm{1}−{z}\right)}\:=\:\frac{\mathrm{1}}{\left.\left(\mathrm{1}−{x}\right)\mathrm{1}−{y}\right)\left(\mathrm{1}−{z}\right)} \\ $$$$\:\:\:{a}+\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\:,\:{b}+\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{1}−{y}}\:\:,\:{c}+\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{1}−{z}} \\ $$$$\left({a}+\mathrm{1}\right)\left({b}+\mathrm{1}\right)\left({c}+\mathrm{1}\right)\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)\left(\mathrm{1}−{z}\right)} \\ $$$$\left({a}+\mathrm{1}\right)\left({b}+\mathrm{1}\right)\left({c}+\mathrm{1}\right)\:=\:{abc} \\ $$$${abc}\:=\:{abc}+{ab}+{bc}+{ca}+{a}+{b}+{c}+\mathrm{1} \\ $$$${ab}+{bc}+{ca}+{a}+{b}+{c}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{2}\left({ab}+{bc}+{ca}\right)+\mathrm{2}\left({a}+{b}+{c}\right)\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ca}\right)+\mathrm{2}\left({a}+{b}+{c}\right)+\mathrm{2}\:=\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} +\mathrm{2}\left({a}+{b}+{c}\right)+\mathrm{2}\:=\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\:\:\left({a}+{b}+{c}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\:=\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\:\:\:\:{now}\:\:\left({a}+{b}+{c}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\:\geqslant\mathrm{1} \\ $$$$\:\:\:\:{so}\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant\mathrm{1} \\ $$$$\:\:\:\:\left(\:\frac{{x}}{\mathrm{1}−{x}}\right)^{\mathrm{2}} +\left(\frac{{y}}{\mathrm{1}−{y}}\right)^{\mathrm{2}} +\left(\frac{{z}}{\mathrm{1}−{z}}\right)^{\mathrm{2}} \geqslant\:\mathrm{1}\: \\ $$

Commented by mr W last updated on 28/Aug/23

great! thanks!

$${great}!\:{thanks}! \\ $$

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