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

AI Thread Summary
The discussion centers on proving the inequality $\dfrac{1}{u-1}+\dfrac{1}{v-1}+\dfrac{1}{x-1}+\dfrac{1}{y-1}>0$ under the conditions that the real numbers $u, v, x, y$ satisfy $|u|>1$, $|v|>1$, $|x|>1$, $|y|>1$, and the equation $u+v+x+y+uv(x+y)+xy(u+v)=0$. Participants explore various mathematical approaches and transformations to manipulate the given conditions and derive the desired inequality. The conversation emphasizes the significance of the constraints on the absolute values of the variables and their implications for the inequality. The challenge is framed as a problem in algebraic manipulation and inequality proof techniques. The thread highlights the complexity of the problem and the collaborative effort to reach a solution.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
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 $\dfrac{1}{u-1}+\dfrac{1}{v-1}+\dfrac{1}{x-1}+\dfrac{1}{y-1}>0$.
 
Mathematics news on Phys.org
anemone said:
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 $\dfrac{1}{u-1}+\dfrac{1}{v-1}+\dfrac{1}{x-1}+\dfrac{1}{y-1}>0---(2)$.
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 $\dfrac {2}{x^2-1}+\dfrac{2}{u^2-1}>0$
 
Albert said:
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,

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.
 
anemone said:
Hi Albert,

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.
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
 
Last edited:
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top