Hoi there,
i'm trying to show something seemingly simple...
$ \left(\sum_{j=k}^n {n\choose j}p^j(1-p)^{n-j}\right)\left(\sum_{j=l}^n {n\choose j}p^j(1-p)^{n-j}\right)\geq \sum_{j=l+k}^n {n\choose j}p^j(1-p)^{n-j}$
Ofcourse $\sum_{j=0}^np^k(1-p)^{n-k}{n\choose j} = 1$ and $p\in (0,1)$
Let's write:
$a= \sum_{j=l+k}^na_j, \ \ \, \ \ \ b= \sum_{j=k}^{l+k-1}a_j, \ \ \ c = \sum_{j=l}^{l+k-1}a_j\\$
$a_j = {n\choose j}p^j(1-p)^{n-j}$
Then I want to show no matter how $l,k\leq n$ are chosen we have$(a+b)(a+c)\geq a$. But I'm 100% stuck on this...the only things we know are $a,b,c\leq 1$ and w.l.o.g. $b\geq c$.
Anyone maybe having an extreme clever insight? Some inequality we should be using?