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Let $n\geq 6$ and $d\geq 3$. Prove that if $l<d+n-1$, then $${l\choose n}+nl-{d+n\choose d}<0.$$ This is the necessary condition in my paper. Experiments show that it is true and $<0$. I tried by expanding but could not find any special for showing that it is $<0$. Thanks for any idea.

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Because$$ \binom{l}{n} < \binom{d + n - 1}{n}, \quad nl < n(d + n - 1), $$ then it suffices to prove that$$ \binom{d + n}{d} \geqslant \binom{d + n - 1}{n} + n(d + n - 1). $$

Since\begin{align*} &\mathrel{\phantom{=}}{} \binom{d + n}{d} - \binom{d + n - 1}{n} = \binom{d + n}{d} - \binom{d + n - 1}{d - 1}\\ &= \binom{d + n - 1}{d} = \frac{1}{d!} (d + n - 1) \cdots (n + 1)n\\ &= \frac{1}{d(d - 1)}·\binom{d + n - 2}{d - 2}·n(d + n - 1), \end{align*} then it suffices to prove that$$ \binom{d + n - 2}{d - 2} \geqslant d(d - 1), $$ which is true in that\begin{align*} \binom{d + n - 2}{d - 2} &\geqslant \binom{d + 5 - 2}{d - 2} = \binom{d + 5 - 2}{5}\\ &= \frac{1}{5!} (d + 3)(d + 2)(d + 1)·d(d - 1)\\ &\geqslant \frac{1}{5!} (3 + 3)(3 + 2)(3 + 1)·d(d - 1) = d(d - 1). \end{align*}

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  • $\begingroup$ It turns out that $n\geqslant5$ suffices. $\endgroup$ – Saad Mar 16 '18 at 9:05

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