# Prove that ${2n \choose n}\left(1-\frac{k}{n}\right)^{k}\leq{2n \choose n+k}$

For $k\leq n$, how do I prove that $${2n \choose n}\left(1-\frac{k}{n}\right)^{k}\leq{2n \choose n+k}\ ?$$

Hint: If $0\lt a\leqslant b$, then $\dfrac{a}b\leqslant\dfrac{a+1}{b+1}$.
Application: The hint yields $\dfrac{n-k}n\leqslant\dfrac{n-k+i}{n+i}$ for every $i\geqslant0$. Multiplying these inequalities for $1\leqslant i\leqslant k$, one gets $$\left(\frac{n-k}n\right)^k\leqslant\frac{n-k+1}{n+1}\frac{n-k+2}{n+2}\cdots\frac{n}{n+k}=\frac{n!\,n!}{(n-k)!(n+k)!}.$$ Hence, $${2n\choose n}\left(\frac{n-k}n\right)^k\leqslant\frac{(2n)!}{n!\,n!}\,\frac{n!\,n!}{(n-k)!(n+k)!}=\frac{(2n)!}{(n-k)!(n+k)!}={2n\choose n+k}.$$