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Question Number 147892 by mathdanisur last updated on 24/Jul/21

if x;y;z>1 then: (√((((x−1)(y−1)(z−1))/((x+1)(y+1)(z+1))) )) < ((xyz)/8)

$${if}\:\:\:{x};{y};{z}>\mathrm{1}\:\:\:{then}: \\ $$ $$\sqrt{\frac{\left({x}−\mathrm{1}\right)\left({y}−\mathrm{1}\right)\left({z}−\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)\left({z}+\mathrm{1}\right)}\:}\:<\:\frac{{xyz}}{\mathrm{8}} \\ $$

Answered by mindispower last updated on 24/Jul/21

((x−1)/(x+1))<(x^2 /4)...? ∀x>1 ⇔4(x−1)<x^2 (x+1) ⇔x^3 +x^2 +4−4x>0 ⇔x(x^2 +1)+4−4x>0 x^2 +1≥2x⇒x(x^2 +1)+4−4x>2x^2 −4x+4=2(x−1)^2 +2>0 ⇒∀x>1 ((x−1)/(x+1))<(x^2 /4),((y−1)/(y+1))<(y^2 /4),((z−1)/(z+1))<(z^2 /4) ⇒(((x−1)(y−1)(z−1))/((x+1)(y+1)(z+1)))<((x^2 y^2 z^2 )/(64)) ⇒(√(((x−1)(y−1)(z−1))/((x+1)(y+1)(z+1)))<((xyz)/8)

$$\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}<\frac{{x}^{\mathrm{2}} }{\mathrm{4}}...?\:\forall{x}>\mathrm{1} \\ $$ $$\Leftrightarrow\mathrm{4}\left({x}−\mathrm{1}\right)<{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right) \\ $$ $$\Leftrightarrow{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{4}−\mathrm{4}{x}>\mathrm{0} \\ $$ $$\Leftrightarrow{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{4}−\mathrm{4}{x}>\mathrm{0} \\ $$ $${x}^{\mathrm{2}} +\mathrm{1}\geqslant\mathrm{2}{x}\Rightarrow{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{4}−\mathrm{4}{x}>\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}=\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}>\mathrm{0} \\ $$ $$\Rightarrow\forall{x}>\mathrm{1}\:\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}<\frac{{x}^{\mathrm{2}} }{\mathrm{4}},\frac{{y}−\mathrm{1}}{{y}+\mathrm{1}}<\frac{{y}^{\mathrm{2}} }{\mathrm{4}},\frac{{z}−\mathrm{1}}{{z}+\mathrm{1}}<\frac{{z}^{\mathrm{2}} }{\mathrm{4}} \\ $$ $$\Rightarrow\frac{\left({x}−\mathrm{1}\right)\left({y}−\mathrm{1}\right)\left({z}−\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)\left({z}+\mathrm{1}\right)}<\frac{{x}^{\mathrm{2}} {y}^{\mathrm{2}} {z}^{\mathrm{2}} }{\mathrm{64}} \\ $$ $$\Rightarrow\sqrt{\frac{\left({x}−\mathrm{1}\right)\left({y}−\mathrm{1}\right)\left({z}−\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)\left({z}+\mathrm{1}\right.}}<\frac{{xyz}}{\mathrm{8}} \\ $$ $$ \\ $$

Commented bymathdanisur last updated on 24/Jul/21

Thankyou Sir cool

$${Thankyou}\:{Sir}\:{cool} \\ $$

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