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Question Number 58789 by ANTARES VY last updated on 30/Apr/19

Commented by ANTARES VY last updated on 30/Apr/19

E=?

$$\mathrm{E}=? \\ $$

Commented by tanmay last updated on 30/Apr/19

E=(1/(x+yz))+(1/(y+xz))+(1/(z+xy)) =(1/(x(x+y+z)+yz))+(1/(y(x+y+z)+xz))+(1/(z(x+y+z)+xy)) =(1/(x^2 +xy+xz+yz))+(1/(xy+y^2 +yz+xz))+(1/(xz+yz+z^2 +xy)) =(1/(x(x+y)+z(x+y)))+(1/(y(x+y)+z(x+y)))+(1/(z(x+z)+y(x+z))) =(1/((x+y)(x+z)))+(1/((x+y)(y+z)))+(1/((x+z)(z+y))) =((y+z+z+x+x+y)/((x+y)(y+z)(z+x))) =((2(x+y+z))/((x+y)(y+z)(z+x))) (x+y+z)^3 =x^3 +y^3 +z^3 +3(x+y)(y+z)(z+x) 1−4=3(x+y)(y+z)(z+x) (x+y)(y+z)(z+x)=−1 hence answer is =((2(x+y+z))/((x+y)(y+z)(z+x))) =((2×1)/(−1))=−2

$${E}=\frac{\mathrm{1}}{{x}+{yz}}+\frac{\mathrm{1}}{{y}+{xz}}+\frac{\mathrm{1}}{{z}+{xy}} \\ $$$$=\frac{\mathrm{1}}{{x}\left({x}+{y}+{z}\right)+{yz}}+\frac{\mathrm{1}}{{y}\left({x}+{y}+{z}\right)+{xz}}+\frac{\mathrm{1}}{{z}\left({x}+{y}+{z}\right)+{xy}} \\ $$$$=\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{xy}+{xz}+{yz}}+\frac{\mathrm{1}}{{xy}+{y}^{\mathrm{2}} +{yz}+{xz}}+\frac{\mathrm{1}}{{xz}+{yz}+{z}^{\mathrm{2}} +{xy}} \\ $$$$=\frac{\mathrm{1}}{{x}\left({x}+{y}\right)+{z}\left({x}+{y}\right)}+\frac{\mathrm{1}}{{y}\left({x}+{y}\right)+{z}\left({x}+{y}\right)}+\frac{\mathrm{1}}{{z}\left({x}+{z}\right)+{y}\left({x}+{z}\right)} \\ $$$$=\frac{\mathrm{1}}{\left({x}+{y}\right)\left({x}+{z}\right)}+\frac{\mathrm{1}}{\left({x}+{y}\right)\left({y}+{z}\right)}+\frac{\mathrm{1}}{\left({x}+{z}\right)\left({z}+{y}\right)} \\ $$$$=\frac{{y}+{z}+{z}+{x}+{x}+{y}}{\left({x}+{y}\right)\left({y}+{z}\right)\left({z}+{x}\right)} \\ $$$$=\frac{\mathrm{2}\left({x}+{y}+{z}\right)}{\left({x}+{y}\right)\left({y}+{z}\right)\left({z}+{x}\right)} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{3}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} +\mathrm{3}\left({x}+{y}\right)\left({y}+{z}\right)\left({z}+{x}\right) \\ $$$$\mathrm{1}−\mathrm{4}=\mathrm{3}\left({x}+{y}\right)\left({y}+{z}\right)\left({z}+{x}\right) \\ $$$$\left({x}+{y}\right)\left({y}+{z}\right)\left({z}+{x}\right)=−\mathrm{1} \\ $$$$\boldsymbol{{hence}}\:\boldsymbol{{answer}}\:\boldsymbol{{is}} \\ $$$$=\frac{\mathrm{2}\left(\boldsymbol{{x}}+\boldsymbol{{y}}+\boldsymbol{{z}}\right)}{\left(\boldsymbol{{x}}+\boldsymbol{{y}}\right)\left(\boldsymbol{{y}}+\boldsymbol{{z}}\right)\left(\boldsymbol{{z}}+\boldsymbol{{x}}\right)} \\ $$$$=\frac{\mathrm{2}×\mathrm{1}}{−\mathrm{1}}=−\mathrm{2} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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