Question Number 1448 by 123456 last updated on 05/Aug/15
![x=(x_1 ,x_2 ),y=(y_1 ,y_2 ) η:[0,1)^4 →[0,1] η(x,y):=med[(((1−x_1 )^y_1 +(1−y_1 )^x_1 )/2),(((1−x_2 )^y_2 +(1−y_2 )^x_2 )/2)] med(x,y):=((min(x,y)+max(x,y))/2) η(x,y)=^? η(y,x) η(x,y)=0⇔^? x=y η(x,z)≤^? η(x,y)+η(y,z)](https://www.tinkutara.com/question/Q1448.png)
$$\boldsymbol{{x}}=\left({x}_{\mathrm{1}} ,{x}_{\mathrm{2}} \right),\boldsymbol{{y}}=\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right) \\ $$$$\eta:\left[\mathrm{0},\mathrm{1}\right)^{\mathrm{4}} \rightarrow\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\eta\left(\boldsymbol{{x}},\boldsymbol{{y}}\right):=\mathrm{med}\left[\frac{\left(\mathrm{1}−{x}_{\mathrm{1}} \right)^{{y}_{\mathrm{1}} } +\left(\mathrm{1}−{y}_{\mathrm{1}} \right)^{{x}_{\mathrm{1}} } }{\mathrm{2}},\frac{\left(\mathrm{1}−{x}_{\mathrm{2}} \right)^{{y}_{\mathrm{2}} } +\left(\mathrm{1}−{y}_{\mathrm{2}} \right)^{{x}_{\mathrm{2}} } }{\mathrm{2}}\right] \\ $$$$\mathrm{med}\left({x},{y}\right):=\frac{\mathrm{min}\left({x},{y}\right)+\mathrm{max}\left({x},{y}\right)}{\mathrm{2}} \\ $$$$\eta\left(\boldsymbol{{x}},\boldsymbol{{y}}\right)\overset{?} {=}\eta\left(\boldsymbol{{y}},\boldsymbol{{x}}\right) \\ $$$$\eta\left(\boldsymbol{{x}},\boldsymbol{{y}}\right)=\mathrm{0}\overset{?} {\Leftrightarrow}\boldsymbol{{x}}=\boldsymbol{{y}} \\ $$$$\eta\left(\boldsymbol{{x}},\boldsymbol{{z}}\right)\overset{?} {\leqslant}\eta\left(\boldsymbol{{x}},\boldsymbol{{y}}\right)+\eta\left(\boldsymbol{{y}},\boldsymbol{{z}}\right) \\ $$
Commented by prakash jain last updated on 05/Aug/15
$$\eta\left(\boldsymbol{{x}},\boldsymbol{{y}}\right)=\eta\left(\boldsymbol{{y}},\boldsymbol{{x}}\right) \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{symmetric}\:\mathrm{by}\:\mathrm{definition}. \\ $$
Commented by Rasheed Soomro last updated on 10/Aug/15
![The definition [in all possible cases x>y , x=y , x<y ] med(x,y):=((min(x,y)+max(x,y))/2) is equaivalent to med(x,y):=((x+y)/2) This will help in calculation of η(x , y) η(x , y)=(((((1−x_1 )^y_1 +(1−y_1 )^x_1 )/2)+(((1−x_2 )^y_2 +(1−y_2 )^x_2 )/2))/2) =(((1−x_1 )^y_1 +(1−y_1 )^x_1 +(1−x_2 )^y_2 +(1−y_2 )^x_2 )/4)........I Now η(y,x)=(((1−y_1 )^x_1 +(1−x_1 )^y_1 +(1−y_2 )^x_2 +(1−x_2 )^y_2 )/4) =(((1−x_1 )^y_1 +(1−y_1 )^x_1 +(1−x_2 )^y_2 +(1−y_2 )^x_2 )/4)........II From I and II η(x,y) =η(y,x) ...........................A From I η(x,y)= (((1−x_1 )^y_1 +(1−y_1 )^x_1 +(1−x_2 )^y_2 +(1−y_2 )^x_2 )/4) Considering x_1 ,x_2 ,y_1 ,y_2 ∈ [0,1) : (1−x_1 )^y_1 has maximum value 1 and if 1 were included in the domain it would have minimum value 0 Hence 0< (1−x_1 )^y_1 ≤1 Similarly 0<(1−y_1 )^x_1 ≤1 0< (1−x_2 )^y_2 ≤1 0<(1−y_2 )^x_2 ≤1 Adding above four inequalities and then dividing by 4 0 < (((1−x_1 )^y_1 +(1−y_1 )^x_1 +(1−x_2 )^y_2 +(1−y_2 )^x_2 )/4)≤1 Or 0< η(x,y)≤1.....................III Using similar logic 0< η(y,z)≤1 0< η(x,z)≤1 From III η(x,y)≠0.......................................B Let x=y i−e (x_1 ,x_2 )=(y_1 ,y_2 ) Replacing y_1 by x_1 and y_2 by x_2 η(x,y)=η(x,x)=(((1−x_1 )^x_1 +(1−x_1 )^x_1 +(1−x_2 )^x_2 +(1−x_2 )^x_2 )/4) =(((1−x_1 )^x_1 +(1−x_2 )^x_2 )/2)≠0 in general. x=y ⇏ η(x,y)=0.......................C Results: η(x,y) =η(y,x) fromA η(x,y)≠0 fromB x=y ⇏ η(x,y)=0 fromC](https://www.tinkutara.com/question/Q1460.png)
$${The}\:{definition}\:\left[\mathrm{in}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{cases}\:\:\mathrm{x}>\mathrm{y}\:,\:\mathrm{x}=\mathrm{y}\:,\:\mathrm{x}<\mathrm{y}\:\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{med}\left(\mathrm{x},\mathrm{y}\right):=\frac{\mathrm{min}\left(\mathrm{x},\mathrm{y}\right)+\mathrm{max}\left(\mathrm{x},\mathrm{y}\right)}{\mathrm{2}} \\ $$$${is}\:{equaivalent}\:{to} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{med}\left(\mathrm{x},\mathrm{y}\right):=\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}} \\ $$$$\mathrm{T}{his}\:{will}\:{help}\:{in}\:{calculation}\:{of}\:\eta\left(\boldsymbol{\mathrm{x}}\:,\:\boldsymbol{\mathrm{y}}\right) \\ $$$$\:\:\:\:\eta\left(\boldsymbol{\mathrm{x}}\:,\:\boldsymbol{\mathrm{y}}\right)=\frac{\frac{\left(\mathrm{1}−\mathrm{x}_{\mathrm{1}} \right)^{\mathrm{y}_{\mathrm{1}} } +\left(\mathrm{1}−\mathrm{y}_{\mathrm{1}} \right)^{\mathrm{x}_{\mathrm{1}} } }{\mathrm{2}}+\frac{\left(\mathrm{1}−\mathrm{x}_{\mathrm{2}} \right)^{\mathrm{y}_{\mathrm{2}} } +\left(\mathrm{1}−\mathrm{y}_{\mathrm{2}} \right)^{\mathrm{x}_{\mathrm{2}} } }{\mathrm{2}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{1}−\mathrm{x}_{\mathrm{1}} \right)^{\mathrm{y}_{\mathrm{1}} } +\left(\mathrm{1}−\mathrm{y}_{\mathrm{1}} \right)^{\mathrm{x}_{\mathrm{1}} } +\left(\mathrm{1}−\mathrm{x}_{\mathrm{2}} \right)^{\mathrm{y}_{\mathrm{2}} } +\left(\mathrm{1}−\mathrm{y}_{\mathrm{2}} \right)^{\mathrm{x}_{\mathrm{2}} } }{\mathrm{4}}……..\mathrm{I} \\ $$$${Now} \\ $$$$\:\:\:\:\:\eta\left(\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{x}}\right)=\frac{\left(\mathrm{1}−\mathrm{y}_{\mathrm{1}} \right)^{\mathrm{x}_{\mathrm{1}} } +\left(\mathrm{1}−\mathrm{x}_{\mathrm{1}} \right)^{\mathrm{y}_{\mathrm{1}} } +\left(\mathrm{1}−\mathrm{y}_{\mathrm{2}} \right)^{\mathrm{x}_{\mathrm{2}} } +\left(\mathrm{1}−\mathrm{x}_{\mathrm{2}} \right)^{\mathrm{y}_{\mathrm{2}} } }{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{1}−\mathrm{x}_{\mathrm{1}} \right)^{\mathrm{y}_{\mathrm{1}} } +\left(\mathrm{1}−\mathrm{y}_{\mathrm{1}} \right)^{\mathrm{x}_{\mathrm{1}} } +\left(\mathrm{1}−\mathrm{x}_{\mathrm{2}} \right)^{\mathrm{y}_{\mathrm{2}} } +\left(\mathrm{1}−\mathrm{y}_{\mathrm{2}} \right)^{\mathrm{x}_{\mathrm{2}} } }{\mathrm{4}}……..\mathrm{II} \\ $$$${From}\:\mathrm{I}\:\:\:\mathrm{and}\:\:\:\mathrm{II} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)\:=\eta\left(\boldsymbol{\mathrm{y}},\boldsymbol{{x}}\right)\:………………………\boldsymbol{\mathrm{A}} \\ $$$${From}\:\boldsymbol{\mathrm{I}} \\ $$$$\:\:\:\:\:\:\:\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)=\:\:\:\frac{\left(\mathrm{1}−\mathrm{x}_{\mathrm{1}} \right)^{\mathrm{y}_{\mathrm{1}} } +\left(\mathrm{1}−\mathrm{y}_{\mathrm{1}} \right)^{\mathrm{x}_{\mathrm{1}} } +\left(\mathrm{1}−\mathrm{x}_{\mathrm{2}} \right)^{\mathrm{y}_{\mathrm{2}} } +\left(\mathrm{1}−\mathrm{y}_{\mathrm{2}} \right)^{\mathrm{x}_{\mathrm{2}} } }{\mathrm{4}} \\ $$$${Considering}\:\:\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} ,\mathrm{y}_{\mathrm{1}} ,\mathrm{y}_{\mathrm{2}} \:\in\:\left[\mathrm{0},\mathrm{1}\right)\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}−\mathrm{x}_{\mathrm{1}} \right)^{\mathrm{y}_{\mathrm{1}} } \:{has}\:{maximum}\:{value}\:\mathrm{1}\:{and}\:{if}\:\mathrm{1}\:{were}\:{included}\:{in}\: \\ $$$${the}\:{domain}\:{it}\:{would}\:{have}\:{minimum}\:{value}\:\mathrm{0} \\ $$$${Hence}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}<\:\left(\mathrm{1}−\mathrm{x}_{\mathrm{1}} \right)^{\mathrm{y}_{\mathrm{1}} } \leqslant\mathrm{1} \\ $$$${Similarly} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}<\left(\mathrm{1}−\mathrm{y}_{\mathrm{1}} \right)^{\mathrm{x}_{\mathrm{1}} } \:\leqslant\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}<\:\left(\mathrm{1}−\mathrm{x}_{\mathrm{2}} \right)^{\mathrm{y}_{\mathrm{2}} } \:\leqslant\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}<\left(\mathrm{1}−\mathrm{y}_{\mathrm{2}} \right)^{\mathrm{x}_{\mathrm{2}} } \:\leqslant\mathrm{1} \\ $$$${Adding}\:{above}\:{four}\:{inequalities}\:{and}\:{then}\:{dividing}\:{by}\:\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:<\:\:\frac{\left(\mathrm{1}−\mathrm{x}_{\mathrm{1}} \right)^{\mathrm{y}_{\mathrm{1}} } +\left(\mathrm{1}−\mathrm{y}_{\mathrm{1}} \right)^{\mathrm{x}_{\mathrm{1}} } +\left(\mathrm{1}−\mathrm{x}_{\mathrm{2}} \right)^{\mathrm{y}_{\mathrm{2}} } +\left(\mathrm{1}−\mathrm{y}_{\mathrm{2}} \right)^{\mathrm{x}_{\mathrm{2}} } }{\mathrm{4}}\leqslant\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:{Or}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}<\:\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)\leqslant\mathrm{1}…………………\boldsymbol{\mathrm{III}} \\ $$$${Using}\:{similar}\:{logic} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}<\:\eta\left(\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{z}}\right)\leqslant\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}<\:\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{z}}\right)\leqslant\mathrm{1} \\ $$$${From}\:\boldsymbol{\mathrm{III}}\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)\neq\mathrm{0}…………………………………\boldsymbol{\mathrm{B}} \\ $$$${Let}\:\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{y}}\:\:\boldsymbol{\mathrm{i}}−\boldsymbol{\mathrm{e}}\:\:\left(\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} \right)=\left(\mathrm{y}_{\mathrm{1}} ,\mathrm{y}_{\mathrm{2}} \right)\: \\ $$$${Replacing}\:\mathrm{y}_{\mathrm{1}} \:{by}\:\mathrm{x}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{y}_{\mathrm{2}} \:{by}\:\mathrm{x}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)=\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{x}}\right)=\frac{\left(\mathrm{1}−\mathrm{x}_{\mathrm{1}} \right)^{\mathrm{x}_{\mathrm{1}} } +\left(\mathrm{1}−\mathrm{x}_{\mathrm{1}} \right)^{\mathrm{x}_{\mathrm{1}} } +\left(\mathrm{1}−\mathrm{x}_{\mathrm{2}} \right)^{\mathrm{x}_{\mathrm{2}} } +\left(\mathrm{1}−\mathrm{x}_{\mathrm{2}} \right)^{\mathrm{x}_{\mathrm{2}} } }{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{1}−\mathrm{x}_{\mathrm{1}} \right)^{\mathrm{x}_{\mathrm{1}} } +\left(\mathrm{1}−\mathrm{x}_{\mathrm{2}} \right)^{\mathrm{x}_{\mathrm{2}} } }{\mathrm{2}}\neq\mathrm{0}\:{in}\:{general}. \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{y}}\:\:\nRightarrow\:\:\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)=\mathrm{0}…………………..\boldsymbol{\mathrm{C}} \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{Results}}:\:\:\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)\:=\eta\left(\boldsymbol{\mathrm{y}},\boldsymbol{{x}}\right)\:\:{from}\boldsymbol{\mathrm{A}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)\neq\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{from}\boldsymbol{\mathrm{B}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{y}}\:\:\nRightarrow\:\:\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)=\mathrm{0}\:\:{from}\boldsymbol{\mathrm{C}} \\ $$
Answered by Rasheed Soomro last updated on 14/Aug/15
$$\mathrm{In}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{cases}\:\left(\:\mathrm{x}<\mathrm{y}\:,\:\mathrm{x}=\mathrm{y}\:,\:\mathrm{x}>\mathrm{y}\:\right)\:\mathrm{the}\:\mathrm{definition}: \\ $$$$\:\:\:\:\:\mathrm{med}\left({x},{y}\right):=\frac{\mathrm{min}\left({x},{y}\right)+\mathrm{max}\left({x},{y}\right)}{\mathrm{2}}\:\:\:\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{med}\left({x},{y}\right):=\frac{{x}+{y}}{\mathrm{2}} \\ $$$$\mathrm{This}\:\mathrm{will}\:\mathrm{help}\:\mathrm{in}\:\mathrm{calculation}\:\mathrm{of}\:\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)=\frac{\frac{\left(\mathrm{1}−{x}_{\mathrm{1}} \right)^{{y}_{\mathrm{1}} } +\left(\mathrm{1}−{y}_{\mathrm{1}} \right)^{{x}_{\mathrm{1}} } }{\mathrm{2}}\:+\frac{\left(\mathrm{1}−{x}_{\mathrm{2}} \right)^{{y}_{\mathrm{2}} } +\left(\mathrm{1}−{y}_{\mathrm{2}} \right)^{{x}_{\mathrm{2}} } }{\mathrm{2}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{1}−{x}_{\mathrm{1}} \right)^{{y}_{\mathrm{1}} } +\left(\mathrm{1}−{y}_{\mathrm{1}} \right)^{{x}_{\mathrm{1}} } +\left(\mathrm{1}−{x}_{\mathrm{2}} \right)^{{y}_{\mathrm{2}} } +\left(\mathrm{1}−{y}_{\mathrm{2}} \right)^{{x}_{\mathrm{2}} } }{\mathrm{4}}……..\left(\boldsymbol{\mathrm{I}}\right) \\ $$$$\:\:\therefore\:\:\:\:\eta\left(\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{x}}\right)\:=\:\frac{\left(\mathrm{1}−{y}_{\mathrm{1}} \right)^{{x}_{\mathrm{1}} } +\left(\mathrm{1}−{x}_{\mathrm{1}} \right)^{{y}_{\mathrm{1}} } +\left(\mathrm{1}−{y}_{\mathrm{2}} \right)^{{x}_{\mathrm{2}} } +\left(\mathrm{1}−{x}_{\mathrm{2}} \right)^{{y}_{\mathrm{2}} } }{\mathrm{4}}……\left(\boldsymbol{\mathrm{II}}\right) \\ $$$$\:\:\boldsymbol{\mathrm{From}}\:\:\:\boldsymbol{\mathrm{I}}\:\:\:\:\mathrm{and}\:\:\:\:\boldsymbol{\mathrm{II}}\:\:\:\::\:\:\:\:\:\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)=\eta\left(\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{x}}\right)……………………..\boldsymbol{\mathrm{A}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ast−\ast−\ast \\ $$$$\boldsymbol{\mathrm{Considering}}\:\:\:{x}_{\mathrm{1}} ,{y}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,{y}_{\mathrm{2}} \:\in\:\left[\mathrm{0},\mathrm{1}\right)\:\boldsymbol{\mathrm{in}}\:\:\:\left(\boldsymbol{\mathrm{I}}\right)\:\:\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{have}}\:, \\ $$$$\:\:\:\:\:\left(\mathrm{1}−{x}_{\mathrm{1}} \right)^{{y}_{\mathrm{1}} } \:\boldsymbol{\mathrm{has}}\:\boldsymbol{\mathrm{maximum}}\:\boldsymbol{\mathrm{value}}\:\:\:\:\mathrm{1}\:\:\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{if}}\:\mathrm{1}\:\boldsymbol{\mathrm{were}}\:\:\boldsymbol{\mathrm{included}}\:\boldsymbol{\mathrm{in}} \\ $$$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{domain}}\:\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{would}}\:\boldsymbol{\mathrm{have}}\:\boldsymbol{\mathrm{minimum}}\:\boldsymbol{\mathrm{value}}\:\:\:\mathrm{0}\: \\ $$$$\mathrm{That}\:\mathrm{is} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}<\left(\mathrm{1}−{x}_{\mathrm{1}} \right)^{{y}_{\mathrm{1}} } \leqslant\mathrm{1} \\ $$$$\mathrm{Similarily}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}<\left(\mathrm{1}−{y}_{\mathrm{1}} \right)^{{x}_{\mathrm{1}} } \leqslant\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{0}<\left(\mathrm{1}−{x}_{\mathrm{2}} \right)^{{y}_{\mathrm{2}} } \leqslant\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{0}<\left(\mathrm{1}−{y}_{\mathrm{2}} \right)^{{x}_{\mathrm{2}} } \leqslant\mathrm{1} \\ $$$$\mathrm{Adding}\:\mathrm{above}\:\mathrm{four}\:\mathrm{inequalities}\:\mathrm{and}\:\mathrm{then}\:\mathrm{dividing}\:\mathrm{by}\:\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{0}<\frac{\left(\mathrm{1}−{x}_{\mathrm{1}} \right)^{{y}_{\mathrm{1}} } +\left(\mathrm{1}−{y}_{\mathrm{1}} \right)^{{x}_{\mathrm{1}} } +\left(\mathrm{1}−{x}_{\mathrm{2}} \right)^{{y}_{\mathrm{2}} } +\left(\mathrm{1}−{y}_{\mathrm{2}} \right)^{{x}_{\mathrm{2}} } }{\mathrm{4}}\leqslant\mathrm{1} \\ $$$$\mathrm{Or}\:\:\:\mathrm{0}<\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)\leqslant\mathrm{1}…………………….\left(\boldsymbol{\mathrm{III}}\right) \\ $$$$\mathrm{Similar}\:\mathrm{logic}\:\mathrm{leads}\:\mathrm{us} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}<\eta\left(\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{z}}\right)\leqslant\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}<\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{z}}\right)\leqslant\mathrm{1} \\ $$$$\mathrm{From}\:\:\:\left(\:\boldsymbol{\mathrm{III}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)\neq\mathrm{0}…………………………\boldsymbol{\mathrm{B}}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ast−\ast−\ast \\ $$$$\:\boldsymbol{\mathrm{Let}}\:\:\:\:\:\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{x}}\right)=\frac{\left(\mathrm{1}−{x}_{\mathrm{1}} \right)^{{x}_{\mathrm{1}} } +\left(\mathrm{1}−{x}_{\mathrm{1}} \right)^{{x}_{\mathrm{1}} } +\left(\mathrm{1}−{x}_{\mathrm{2}} \right)^{{x}_{\mathrm{2}} } +\left(\mathrm{1}−{x}_{\mathrm{2}} \right)^{{x}_{\mathrm{2}} } }{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{1}−{x}_{\mathrm{1}} \right)^{{x}_{\mathrm{1}} } +\left(\mathrm{1}−{x}_{\mathrm{2}} \right)^{{x}_{\mathrm{2}} } }{\mathrm{2}}\neq\mathrm{0}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{general}}. \\ $$$$\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{y}}\nRightarrow\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)=\mathrm{0}……………………………….\boldsymbol{\mathrm{C}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ast−\ast−\ast \\ $$$$\boldsymbol{\mathrm{Results}}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)=\eta\left(\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{x}}\right)\:\:\:\:\:\:\mathrm{From}\:\boldsymbol{\mathrm{A}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)\neq\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{F}}{rom}\:\boldsymbol{\mathrm{B}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{y}}\nRightarrow\eta\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)=\mathrm{0}\:\:\:\:\mathrm{F}{rom}\:\boldsymbol{\mathrm{C}} \\ $$$$\boldsymbol{\mathrm{Note}}:\:\boldsymbol{\mathrm{This}}\:\:\boldsymbol{\mathrm{answer}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{silent}}\:\boldsymbol{\mathrm{about}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\eta}\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{z}}\right)\overset{?} {\leqslant}\boldsymbol{\eta}\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)\:+\boldsymbol{\eta}\left(\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{z}}\right) \\ $$$$ \\ $$