MHB Is There a Solution to the Challenge of Inequality?

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The discussion centers around a mathematical inequality involving the variables k, l, m, and n, constrained between 0 and 1, and their product equating to the product of their complements. A mistake was acknowledged regarding a minus sign that was overlooked in the initial formulation. The participants express appreciation for the valid proof provided by lfdahl, highlighting its uniqueness compared to existing proofs. The conversation emphasizes collaboration and sharing knowledge within the community. Overall, the thread showcases a constructive exchange focused on resolving a mathematical challenge related to inequality.
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Given that $0<k,\,l,\,m,\,n<1$ and $klmn=(1-k)(1-l)(1-m)(1-n)$, show that $(k+l+m+n)-(k+m)(l+n)\ge1$.
 
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I´m so sorry for my mistake: A minus-sign was overlooked, and the coupling of the four variables k, l, m, n is not obvious. Sorry!

Given:
\[klmn = (1-k)(1-l)(1-m)(1-n),\: \: \: \: 0 < k,l,m,n < 1\]

This equality is only possible, if LHS and RHS have equal factors* (not necessarily in the same order).

\[klmn = (1-k)(1-l)(1-m)(1-n)\\\\ =1 - (k+l+m+n)+(k+m)(l+n)+mk+nl-(klm+kln+kmn+lmn)+klmn\\\\ \Rightarrow (k+l+m+n)-(k+m)(l+n)=1+mk(1-n-l)+nl(1-m-k)\]

Using (*): WLOG I can take $k = 1-l$, and $n = 1-m$. Thus, the variables are coupled pairwise. I could as well take $k = 1-n$ and $l = 1-m$. The outcome will be exactly the same. But taking $k = 1-m$ and $l = 1-n$ doesn´t work. I don´t know why ...
(- well, it works, because the $\ge$ is still valid).\[(k+l+m+n)-(k+m)(l+n)=1+mk((1-l)-n)+nl(1-m-k)\\\\ =1+mk(k-(1-m))+(1-m)(1-k)(1-m-k)\\\\ =1-mk(1-m-k)+(1-m)(1-k)(1-m-k)\\\\ =1+(1-m-k)(1-m-k+mk-mk)\\\\ =1+(1-m-k)^2 \geq 1\]
 
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lfdahl said:
I´m so sorry for my mistake: A minus-sign was overlooked, and the coupling of the four variables k, l, m, n is not obvious. Sorry!

Hey lfdahl, seriously, there's no need to apologize!(Smile) And everything is fine and perfect with your valid proof!:cool:

lfdahl said:
Given:
\[klmn = (1-k)(1-l)(1-m)(1-n),\: \: \: \: 0 < k,l,m,n < 1\]

This equality is only possible, if LHS and RHS have equal factors* (not necessarily in the same order).

\[klmn = (1-k)(1-l)(1-m)(1-n)\\\\ =1 - (k+l+m+n)+(k+m)(l+n)+mk+nl-(klm+kln+kmn+lmn)+klmn\\\\ \Rightarrow (k+l+m+n)-(k+m)(l+n)=1+mk(1-n-l)+nl(1-m-k)\]

Using (*): WLOG I can take $k = 1-l$, and $n = 1-m$. Thus, the variables are coupled pairwise. I could as well take $k = 1-n$ and $l = 1-m$. The outcome will be exactly the same. But taking $k = 1-m$ and $l = 1-n$ doesn´t work. I don´t know why ...
(- well, it works, because the $\ge$ is still valid).\[(k+l+m+n)-(k+m)(l+n)=1+mk((1-l)-n)+nl(1-m-k)\\\\ =1+mk(k-(1-m))+(1-m)(1-k)(1-m-k)\\\\ =1-mk(1-m-k)+(1-m)(1-k)(1-m-k)\\\\ =1+(1-m-k)(1-m-k+mk-mk)\\\\ =1+(1-m-k)^2 \geq 1\]

Well done, lfdahl! Your proof is different from the one that I saw online somewhere, hence, I will post it here to share with the community:

We're given $klmn=(1-k)(1-l)(1-m)(1-n)$, we can rewrite it as $\dfrac{km}{(1-k)(1-m)}=\dfrac{(1-l)(1-n)}{ln}$ and this is also equivalent to $\dfrac{(k+m)-1}{(1-k)(1-m)}=\dfrac{1-(l+n)}{ln}$.

Recall that if we have $a=b$, then $ab= a^2=b^2\ge0$.

Hence, $\dfrac{(k+m)-1}{(1-k)(1-m)}\cdot\dfrac{1-(l+n)}{ln}\ge 0$.

Since $0<k,\,l,\,m,\,n<1$, we see that the product of the terms in the denominator greater than zero, this yields:

$((k+m)-1)(1-(l+n))\ge0$

$k+m-(k+m)(l+n)-1+l+n\ge0$

$k+m+l+n-(k+m)(l+n)\ge1$ (Q.E.D.)
 
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