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anemone
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Given that $0<k,\,l,\,m,\,n<1$ and $klmn=(1-k)(1-l)(1-m)(1-n)$, show that $(k+l+m+n)-(k+m)(l+n)\ge1$.
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!
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).\[(k+l+m+n)-(k+m)(l+n)=1+mk((1-l)-n)+nl(1-m-k)\\\\ =1+mk(k-(1-m))+(1-m)(1-k)(1-m-k)\\\\ =1-mk(1-m-k)+(1-m)(1-k)(1-m-k)\\\\ =1+(1-m-k)(1-m-k+mk-mk)\\\\ =1+(1-m-k)^2 \geq 1\]
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