Binomial inequality problem ${k+n-1 \choose k}\times{k+n+1 \choose k} \leq{k+n \choose k}^2$

Question

Can anyone help we with this problem:
Let $a_n={k+n \choose k} $
Prove that $a_{k-1}a_{k+1}\leq a_k^2 $($\forall k$)
My first idea was using mathematical induction to proof that for every k element of natural number. The basis is easy, but the problem is inductive step.

Answer 1

Simply write down \begin{align} \binom{k+n-1}{k}\binom{k+n+1}{k} = & \frac{(k+n-1)!(k+n+1)!}{(k!)^2(n-1)!(n+1)!}\\ = & \binom{k+n}{k}^2\frac{(k+n+1)(n-1)}{(k+n)(n+1)} \end{align} and show that the fraction we have in the last term is smaller than $1$ (which I'll let you do).