Question Number 15721 by ajfour last updated on 13/Jun/17
$${If}\:\:\:\:\boldsymbol{{x}}+\boldsymbol{{y}}+\boldsymbol{{z}}=\mathrm{1} \\ $$$$\:\:\:\:\:{with}\:\mathrm{0}<{x},\:{y},\:{z}\:<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:{then}\:{find}\:{tbe}\:\:{range}\:{of}\:{values}\:{of} \\ $$$$\:\:\:\frac{\mathrm{1}}{\boldsymbol{{x}}+\boldsymbol{{y}}}+\frac{\mathrm{1}}{\boldsymbol{{y}}+\boldsymbol{{z}}}+\frac{\mathrm{1}}{\boldsymbol{{z}}+\boldsymbol{{x}}}\:. \\ $$
Answered by mrW1 last updated on 13/Jun/17
![(1/(x+y))+(1/(y+z))+(1/(z+x))=(1/(x+y))+(1/(1−x))+(1/(1−y)) let f(x,y)=(1/(x+y))+(1/(1−x))+(1/(1−y)) (∂f/∂x)=−(1/((x+y)^2 ))+(1/((1−x)^2 )) (∂^2 f/∂x^2 )=(2/((x+y)^3 ))+(2/((1−x)^3 )) (∂f/∂y)=−(1/((x+y)^2 ))+(1/((1−y)^2 )) (∂^2 f/∂y^2 )=(2/((x+y)^3 ))+(2/((1−y)^3 )) (∂^2 f/(∂x∂y))=(2/((x+y)^3 )) (∂f/∂x)=−(1/((x+y)^2 ))+(1/((1−x)^2 ))=0 ⇒x+y=1−x ...(i) (∂f/∂y)=−(1/((x+y)^2 ))+(1/((1−y)^2 ))=0 ⇒x+y=1−y ...(ii) from (i)and (ii) ⇒x=y=(1/3) f((1/3),(1/3))=(1/((1/3)+(1/3)))+(1/(1−(1/3)))+(1/(1−(1/3)))=(9/2) ((∂^2 f((1/3),(1/3)))/∂x^2 )=(2/(((2/3))^3 ))+(2/(((2/3))^3 ))=((27)/2)>0 ((∂^2 f((1/3),(1/3)))/∂y^2 )=(2/(((2/3))^3 ))+(2/(((2/3))^3 ))=((27)/2)>0 ((∂^2 f((1/3),(1/3)))/(∂x∂y))=(2/(((2/3))^3 ))=((27)/4)>0 ⇒f((1/3),(1/3))=(9/2) is the minimum! lim_(x→0,y→0) f(x,y) =lim_(x→0) [lim_(y→0) f(x,y)] =lim_(x→0) [lim_(y→0) ((1/(x+y))+(1/(1−x))+(1/(1−y)))] =lim_(x→0) [(1/x)+(1/(1−x))+1] =lim_(x→0) [(1/x)]+1+1 =+∞ ⇒ (1/(x+y))+(1/(y+z))+(1/(z+x)) ∈[(9/2),+∞)](https://www.tinkutara.com/question/Q15727.png)
$$\:\frac{\mathrm{1}}{\boldsymbol{{x}}+\boldsymbol{{y}}}+\frac{\mathrm{1}}{\boldsymbol{{y}}+\boldsymbol{{z}}}+\frac{\mathrm{1}}{\boldsymbol{{z}}+\boldsymbol{{x}}}=\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{y}} \\ $$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{y}} \\ $$$$\frac{\partial\mathrm{f}}{\partial\mathrm{x}}=−\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$\frac{\partial^{\mathrm{2}} \mathrm{f}}{\partial\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{2}}{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{3}} }+\frac{\mathrm{2}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{3}} } \\ $$$$ \\ $$$$\frac{\partial\mathrm{f}}{\partial\mathrm{y}}=−\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{y}\right)^{\mathrm{2}} } \\ $$$$\frac{\partial^{\mathrm{2}} \mathrm{f}}{\partial\mathrm{y}^{\mathrm{2}} }=\frac{\mathrm{2}}{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{3}} }+\frac{\mathrm{2}}{\left(\mathrm{1}−\mathrm{y}\right)^{\mathrm{3}} } \\ $$$$\frac{\partial^{\mathrm{2}} \mathrm{f}}{\partial\mathrm{x}\partial\mathrm{y}}=\frac{\mathrm{2}}{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{3}} } \\ $$$$ \\ $$$$\frac{\partial\mathrm{f}}{\partial\mathrm{x}}=−\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}+\mathrm{y}=\mathrm{1}−\mathrm{x}\:\:\:\:\:…\left(\mathrm{i}\right) \\ $$$$ \\ $$$$\frac{\partial\mathrm{f}}{\partial\mathrm{y}}=−\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{y}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}+\mathrm{y}=\mathrm{1}−\mathrm{y}\:\:\:\:…\left(\mathrm{ii}\right) \\ $$$$ \\ $$$$\mathrm{from}\:\left(\mathrm{i}\right)\mathrm{and}\:\left(\mathrm{ii}\right) \\ $$$$\Rightarrow\mathrm{x}=\mathrm{y}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\frac{\partial^{\mathrm{2}} \mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)}{\partial\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{2}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }+\frac{\mathrm{2}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }=\frac{\mathrm{27}}{\mathrm{2}}>\mathrm{0} \\ $$$$\frac{\partial^{\mathrm{2}} \mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)}{\partial\mathrm{y}^{\mathrm{2}} }=\frac{\mathrm{2}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }+\frac{\mathrm{2}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }=\frac{\mathrm{27}}{\mathrm{2}}>\mathrm{0} \\ $$$$\frac{\partial^{\mathrm{2}} \mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)}{\partial\mathrm{x}\partial\mathrm{y}}=\frac{\mathrm{2}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }=\frac{\mathrm{27}}{\mathrm{4}}>\mathrm{0} \\ $$$$\Rightarrow\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{9}}{\mathrm{2}}\:\mathrm{is}\:\mathrm{the}\:\mathrm{minimum}! \\ $$$$ \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0},\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}f}\left(\mathrm{x},\mathrm{y}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}f}\left(\mathrm{x},\mathrm{y}\right)\right] \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{y}}\right)\right] \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}+\mathrm{1}\right] \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{\mathrm{x}}\right]+\mathrm{1}+\mathrm{1} \\ $$$$=+\infty \\ $$$$ \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}}{\boldsymbol{{x}}+\boldsymbol{{y}}}+\frac{\mathrm{1}}{\boldsymbol{{y}}+\boldsymbol{{z}}}+\frac{\mathrm{1}}{\boldsymbol{{z}}+\boldsymbol{{x}}}\:\in\left[\frac{\mathrm{9}}{\mathrm{2}},+\infty\right) \\ $$
Commented by ajfour last updated on 13/Jun/17
$${thanks}\:{sir},\:{you}\:{are}\:{far}\:{too} \\ $$$${generous}! \\ $$
Commented by mrW1 last updated on 13/Jun/17
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}!\:\mathrm{i}\:\mathrm{do}\:\mathrm{also}\:\mathrm{learn}\:\mathrm{from}\:\mathrm{you}. \\ $$