An Inequality with Constraint X

Problem

Leo Giugiuc has kindly reposted a problem from an artofproblemsolving forum at the CutTheKnotMath facebook page, along with a solution (Proof 1) by Leo Giugiuc and Dan Sitaru.

Given xypq=(1-x)(1-y)(1-p)(1-q), all between 0 and 1. Prove that (x+y+p+q)-(x+y)(p+q) is not less than 1

Proof 1

It is obvious that $\displaystyle x=\frac{1}{1+a},\;$ $\displaystyle y=\frac{1}{1+b},\;$ $\displaystyle p=\frac{1}{1+c},\;$ $\displaystyle q=\frac{1}{1+d},\;$ $a,b,c,d\gt 0\;$ and $abcd=1.$

We need to prove $(x+y-1)(p+q-1)\le 0.\;$ This is equivalent to

$\displaystyle\left(\frac{1-a}{1+a}+\frac{1-b}{1+b}\right)\left(\frac{1-c}{1+c}+\frac{1-d}{1+d}\right)\le 0.$

In case of equality there is nothing else to prove. So, let's assume that neither factor is zero: one is negative, the other is positive. Suppose, WLOG, that $\displaystyle\left(\frac{1-a}{1+a}+\frac{1-b}{1+b}\right)\lt 0\;$ and $\displaystyle\left(\frac{1-c}{1+c}+\frac{1-d}{1+d}\right)\gt 0.\;$ Under these assumptions,

$\displaystyle\left(\frac{1-a}{1+a}+\frac{1-b}{1+b}\right)\lt \frac{1-a}{1+a}+\frac{1-\displaystyle\frac{1}{a}}{1+\displaystyle\frac{1}{a}},$

Thus, $\displaystyle\frac{1-b}{1+b}\lt\frac{1-\displaystyle\frac{1}{a}}{1+\displaystyle\frac{1}{a}},$

implying $\displaystyle b\gt\frac{1}{a},\;$ because function $\displaystyle f(t)=\frac{1-t}{1+t}\;$ is strictly increasing on $(0,\infty ).\;$ It follows that $ab\gt 1\;$ and, since $abcd=1,\;$ $cd\lt 1.$

But $\displaystyle d\lt\frac{1}{c}\;$ implies

$\displaystyle\frac{1-c}{1+c}+\frac{1-d}{1+d}\gt\frac{1-c}{1+c}+\frac{1-\displaystyle\frac{1}{c}}{1+\displaystyle\frac{1}{c}}=0,$

as expected.

Proof 2

With the same substitution, $\displaystyle x=\frac{1}{1+a},\;$ $\displaystyle y=\frac{1}{1+b},\;$ $\displaystyle p=\frac{1}{1+c},\;$ $\displaystyle q=\frac{1}{1+d},\;$ with $a,b,c,d\gt 0\;$ and $abcd=1.$

We need to prove $(x+y-1)(p+q-1)\le 0,\;$ or, $(1-ab)(1-cd)\le 0.\;$ But this is obvious, since $(ab)(cd)=1.$

 

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