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Question Number 15721 by ajfour last updated on 13/Jun/17

If    x+y+z=1       with 0<x, y, z <(1/2)   then find tbe  range of values of     (1/(x+y))+(1/(y+z))+(1/(z+x)) .

$${If}\:\:\:\:\boldsymbol{{x}}+\boldsymbol{{y}}+\boldsymbol{{z}}=\mathrm{1} \\ $$ $$\:\:\:\:\:{with}\:\mathrm{0}<{x},\:{y},\:{z}\:<\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$\:{then}\:{find}\:{tbe}\:\:{range}\:{of}\:{values}\:{of} \\ $$ $$\:\:\:\frac{\mathrm{1}}{\boldsymbol{{x}}+\boldsymbol{{y}}}+\frac{\mathrm{1}}{\boldsymbol{{y}}+\boldsymbol{{z}}}+\frac{\mathrm{1}}{\boldsymbol{{z}}+\boldsymbol{{x}}}\:. \\ $$

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),+∞)

$$\:\frac{\mathrm{1}}{\boldsymbol{{x}}+\boldsymbol{{y}}}+\frac{\mathrm{1}}{\boldsymbol{{y}}+\boldsymbol{{z}}}+\frac{\mathrm{1}}{\boldsymbol{{z}}+\boldsymbol{{x}}}=\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{y}} \\ $$ $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{y}} \\ $$ $$\frac{\partial\mathrm{f}}{\partial\mathrm{x}}=−\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} } \\ $$ $$\frac{\partial^{\mathrm{2}} \mathrm{f}}{\partial\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{2}}{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{3}} }+\frac{\mathrm{2}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{3}} } \\ $$ $$ \\ $$ $$\frac{\partial\mathrm{f}}{\partial\mathrm{y}}=−\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{y}\right)^{\mathrm{2}} } \\ $$ $$\frac{\partial^{\mathrm{2}} \mathrm{f}}{\partial\mathrm{y}^{\mathrm{2}} }=\frac{\mathrm{2}}{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{3}} }+\frac{\mathrm{2}}{\left(\mathrm{1}−\mathrm{y}\right)^{\mathrm{3}} } \\ $$ $$\frac{\partial^{\mathrm{2}} \mathrm{f}}{\partial\mathrm{x}\partial\mathrm{y}}=\frac{\mathrm{2}}{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{3}} } \\ $$ $$ \\ $$ $$\frac{\partial\mathrm{f}}{\partial\mathrm{x}}=−\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$ $$\Rightarrow\mathrm{x}+\mathrm{y}=\mathrm{1}−\mathrm{x}\:\:\:\:\:...\left(\mathrm{i}\right) \\ $$ $$ \\ $$ $$\frac{\partial\mathrm{f}}{\partial\mathrm{y}}=−\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{y}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$ $$\Rightarrow\mathrm{x}+\mathrm{y}=\mathrm{1}−\mathrm{y}\:\:\:\:...\left(\mathrm{ii}\right) \\ $$ $$ \\ $$ $$\mathrm{from}\:\left(\mathrm{i}\right)\mathrm{and}\:\left(\mathrm{ii}\right) \\ $$ $$\Rightarrow\mathrm{x}=\mathrm{y}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$ $$\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$ $$\frac{\partial^{\mathrm{2}} \mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)}{\partial\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{2}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }+\frac{\mathrm{2}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }=\frac{\mathrm{27}}{\mathrm{2}}>\mathrm{0} \\ $$ $$\frac{\partial^{\mathrm{2}} \mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)}{\partial\mathrm{y}^{\mathrm{2}} }=\frac{\mathrm{2}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }+\frac{\mathrm{2}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }=\frac{\mathrm{27}}{\mathrm{2}}>\mathrm{0} \\ $$ $$\frac{\partial^{\mathrm{2}} \mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)}{\partial\mathrm{x}\partial\mathrm{y}}=\frac{\mathrm{2}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }=\frac{\mathrm{27}}{\mathrm{4}}>\mathrm{0} \\ $$ $$\Rightarrow\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{9}}{\mathrm{2}}\:\mathrm{is}\:\mathrm{the}\:\mathrm{minimum}! \\ $$ $$ \\ $$ $$\underset{\mathrm{x}\rightarrow\mathrm{0},\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}f}\left(\mathrm{x},\mathrm{y}\right) \\ $$ $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}f}\left(\mathrm{x},\mathrm{y}\right)\right] \\ $$ $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{y}}\right)\right] \\ $$ $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}+\mathrm{1}\right] \\ $$ $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{\mathrm{x}}\right]+\mathrm{1}+\mathrm{1} \\ $$ $$=+\infty \\ $$ $$ \\ $$ $$\Rightarrow\:\:\frac{\mathrm{1}}{\boldsymbol{{x}}+\boldsymbol{{y}}}+\frac{\mathrm{1}}{\boldsymbol{{y}}+\boldsymbol{{z}}}+\frac{\mathrm{1}}{\boldsymbol{{z}}+\boldsymbol{{x}}}\:\in\left[\frac{\mathrm{9}}{\mathrm{2}},+\infty\right) \\ $$

Commented byajfour last updated on 13/Jun/17

thanks sir, you are far too  generous!

$${thanks}\:{sir},\:{you}\:{are}\:{far}\:{too} \\ $$ $${generous}! \\ $$

Commented bymrW1 last updated on 13/Jun/17

thank you sir! i do also learn from you.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}!\:\mathrm{i}\:\mathrm{do}\:\mathrm{also}\:\mathrm{learn}\:\mathrm{from}\:\mathrm{you}. \\ $$

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