Let $0\le k \le n$. Prove that
$$
n\binom{n}{k}\int\_{0}^{\frac{k}{n+1}}t^k(1-t)^{n-k}\,dt \le 1/2.
$$
As far as I know
1) it is proved for $\frac{k}{n+1}\le 1/2$ and 
2) not proved for $1/2 <\frac{k}{n+1}< 1$.
Is it really so?



Estimate of incomplete binomial integral
The following stronger inequality holds for all $k=0,\dots,n$: 
$$
(n+1)\binom{n}{k}\int\_{0}^{\frac{k}{n+1}}t^k(1-t)^{n-k}\,dt \le 1/2.
$$
Indeed, the latter inequality means precisely that the median $m$ of the Beta distribution with parameters $a:=k+1\ge1$ and $b:=n-k+1\ge1$ is no less than $\frac{k}{n+1}=\frac{a-1}{a+b-1}$. 
By the well-known "Mode, Median, and Mean Inequalities" for the Beta distribution (see e.g. page 2 in <http://arxiv.org/abs/1111.0433v1>), 
\begin{equation}
 m\ge\frac{a-1}{a+b-2}\bigwedge\frac{a}{a+b} \ge\frac{a-1}{a+b-1}=\frac{k}{n+1}, 
\end{equation}
as desired. 


