Proving some inequalities with factorials

Question

I'm having some difficulty proving a few inequalities with factorials, and I was hoping someone could point me in the right direction.

The first one is for $k\in\mathbb{N}$ and $0\leq t\leq 1$, that $(k+t)(k+t-1)\cdots(1+t)\leq(k+1)!$. My "proof" so far for this one is since $k+t\leq k+1, k+t-1\leq k, \dots 1+t\leq 2$, it shows that inequality.

Another is that for $n\in\mathbb{N}, k=0,\dots,n-1$ and $0\leq t\leq1$, $(2-t)\cdots(n-k-t)\leq(n-k)!$. This one I am struggling with as I'm not sure what the "$\dots$" means here, ie. what is between those terms.

The final one is for the same $n$ and $k$ as above that $(k+1)!(n-k)!\leq n!$.

Answer 1

Your proof of the first one is correct.

The $\ldots$ in the second one means $(2-t)(3-t)(4-t) \ldots (n - (k-1) - t)(n - k - t)$, and the method to prove it is very similar to the first one... good luck!

@Peter's comment explains a method to solve the final one, if you're also stuck on that:

$$ (k+1)! (n-k)! \leq n! \implies (k+1)! \leq {n \choose k} $$

Answer 2

Your first part is fine - if one adds the remark that all factors $k+t$ are positive.

The second is presumably about $(2-t)(3-t)(4-t)\cdot(n-k-t)$ and essentially the same after substituting $1-t$ for $t$ (so that we want to show $(1+t)(2+t)\cdot(n- k-1+t)\le (n-k)!$, which is obtained from the first case with $n-k-1$ in place of $n$).

Finally, consider the non-negative integer ${n+1\choose k+1}=\frac{(n+1)!}{(k+1)!(n-k)!}=(n+1)\cdot \frac{n!}{(k+1)!(n-k)!}$. From its combinatorical interpretation of choosing $k+1$ out of $n+1$ objects, this integer is $\ge n+1$ if $0\le k\le n$.