Inequality Challenge: Prove $1/(u-1)+1/(v-1)+1/(x-1)+1/(y-1)>0$

[anemone]

Real numbers $u,\,v,\,x,\,y$ satisfy the following conditions:

$|u|>1$, $|v|>1$, $|x|>1$, $|y|>1$, and

$u+v+x+y+uv(x+y)+xy(u+v)=0$

Prove that $\dfrac{1}{u-1}+\dfrac{1}{v-1}+\dfrac{1}{x-1}+\dfrac{1}{y-1}>0$.

 

[Albert1]

Real numbers $u,\,v,\,x,\,y$ satisfy the following conditions:

$|u|>1$, $|v|>1$, $|x|>1$, $|y|>1$, and

$u+v+x+y+uv(x+y)+xy(u+v)=0---(1)$

Prove that $\dfrac{1}{u-1}+\dfrac{1}{v-1}+\dfrac{1}{x-1}+\dfrac{1}{y-1}>0---(2)$.

Spoiler

let :$xy+1=m\neq 0,\,\,uv+1=n\neq 0$
and $x+y=X,\,\, u+v=Y$
then (1) becomes $nX+mY=0---(3)$
we can consider (3) as a line passing through origin $(0,0)=(X,Y)=(x+y,u+v)$
and (2) becomes $\dfrac {2}{x^2-1}+\dfrac{2}{u^2-1}>0$

 

[anemone]

Spoiler

let :$xy+1=m\neq 0,\,\,uv+1=n\neq 0$
and $x+y=X,\,\, u+v=Y$
then (1) becomes $nX+mY=0---(3)$
we can consider (3) as a line passing through origin $(0,0)=(X,Y)=(x+y,u+v)$
and (2) becomes $\dfrac {2}{x^2-1}+\dfrac{2}{u^2-1}>0$

Hi Albert,

Spoiler

I thank you for your constant support in my challenge problems and I appreciate that 100%.

But, your solution to this particular challenge, I am sorry, I just could not follow it...do you mean to say $X=Y=0$, when $X=x+y$ and both $|x|$ and $|y|$ are greater than zero?

Perhaps you have your point there, but the rename of the variables, e.g. the one that uses $X$ to represent $x+y$ is quite confusing to me.

 

[Albert1]

Hi Albert,

Spoiler

I thank you for your constant support in my challenge problems and I appreciate that 100%.

But, your solution to this particular challenge, I am sorry, I just could not follow it...do you mean to say $X=Y=0$, when $X=x+y$ and both $|x|$ and $|y|$ are greater than zero?

Perhaps you have your point there, but the rename of the variables, e.g. the one that uses $X$ to represent $x+y$ is quite confusing to me.

Spoiler

the linear expression of a line nX+mY+c=0 ,if passes through the origin then c=0 )
if x=3 then y=-3 ,and X=3+(-3)=0 ,u=5 ,v=-5 ,Y=5+(-5)=0
I mean X=x+y, and Y=u+v