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I supect that for all $n>k>0$:

$k^2\left\{ \begin{array}{c}n\\k\end{array} \right\}^2 +2k\left\{ \begin{array}{c}n\\k\end{array} \right\}\left\{ \begin{array}{c}n\\k-1\end{array} \right\}+\left\{ \begin{array}{c}n\\k-1\end{array} \right\}^2-(k-1)(k+1)\left\{ \begin{array}{c}n\\k-1\end{array} \right\}\left\{ \begin{array}{c}n\\k+1\end{array} \right\}-(k-1)\left\{ \begin{array}{c}n\\k-1\end{array} \right\}\left\{ \begin{array}{c}n\\k\end{array} \right\}-(k+1)\left\{ \begin{array}{c}n\\k-2\end{array} \right\}\left\{ \begin{array}{c}n\\k+1\end{array} \right\}-\left\{ \begin{array}{c}n\\k-2\end{array} \right\}\left\{ \begin{array}{c}n\\k\end{array} \right\}>0$

where $\left\{ \begin{array}{c}n\\k\end{array} \right\}$ is Stirling's number of the second kind.. and I want to prove it.. I want to do it as simple as possible, but there are too many words in this inequality.. Is there any chance to group the words and see that the left side is the sum of non negative expressions?

I can use: $\left\{ \begin{array}{c}n\\k-1\end{array} \right\}\left\{ \begin{array}{c}n\\k+1\end{array} \right\}<\left\{ \begin{array}{c}n\\k\end{array} \right\}^2$ but this is an induction step so I don't know if $\left\{ \begin{array}{c}n+1\\k-1\end{array} \right\}\left\{ \begin{array}{c}n+1\\k+1\end{array} \right\}<\left\{ \begin{array}{c}n+1\\k\end{array} \right\}^2$

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@Brian M. Scott, exactly.. –  xan Mar 10 '12 at 19:45
    
@BrianM.Scott: Oh... Of course if it's the thing to prove I'll remove my previous comment... I thought it was the next step using the previous result not the same... –  Raymond Manzoni Mar 10 '12 at 19:48
    
I've just made it simpler to form: $(k+1)\left\{ \begin{array}{c}n\\k\end{array} \right\}\left\{ \begin{array}{c}n\\k-1\end{array} \right\} -(k+1)\left\{ \begin{array}{c}n\\k-2\end{array} \right\}\left\{ \begin{array}{c}n\\k+1\end{array} \right\}+\left\{ \begin{array}{c}n\\k-1\end{array} \right\}\left\{ \begin{array}{c}n\\k+1\end{array} \right\}>0$, but I worry this is not the end.. –  xan Mar 10 '12 at 21:26

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