Is There a Solution to the Challenge of Inequality?

[anemone]

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]

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!

Spoiler

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).\[(k+l+m+n)-(k+m)(l+n)=1+mk((1-l)-n)+nl(1-m-k)\\\\ =1+mk(k-(1-m))+(1-m)(1-k)(1-m-k)\\\\ =1-mk(1-m-k)+(1-m)(1-k)(1-m-k)\\\\ =1+(1-m-k)(1-m-k+mk-mk)\\\\ =1+(1-m-k)^2 \geq 1\]

 

[anemone]

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!

Spoiler

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:

Spoiler

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.)