Let $B(a,b;x):=\int_0^x t^{a-1}(1-t)^{b-1}dt$ denote, as usual, the value of the incomplete Beta function. We have to show that 
$$
n\binom{n}{k}B(k+1,n-k+1,\tfrac k{n+1}) \le 1/2 \quad\text{for $k=0,\dots,n$}. \tag{1}
$$  
As suggested by Brendan McKay, it should be easy to use asymptotic analysis to check that $(1)$ holds for any real $c>1/2$ if $n$ is large enough. Thus, we can use the reverse induction in $n$. That is, it is enough to show that, if $(1)$ holds for some natural $n+1\ge2$ in place of $n$ and some real $c>0$ in place of $1/2$, then it holds as written, but with the same real $c>0$ in place of $1/2$. The key here is the simple identity 
\begin{equation}
	B(a,b;x)=B(a+1,b;x)+B(a,b+1;x),   
\end{equation}
which yields, by the reverse induction, 
$$
	B(k+1,n-k+1,\tfrac k{n+1})=B(k+2,n-k+1,\tfrac k{n+1})+B(k+1,n-k+2,\tfrac k{n+1})
$$
$$	
	=B(k+2,n-k+1,\tfrac{k+1}{n+2})+B(k+1,n-k+2,\tfrac k{n+2})+I_1-I_2
$$
$$	
	\le\frac c{n+1}/\binom{n+1}{k+1} +\frac c{n+1}/\binom{n+1}{k}+I_1-I_2 
	=\frac{c(n+2)n}{(n+1)^2}\frac1{n\binom nk}+I_1-I_2 
	<\frac c{n\binom nk}+I_1-I_2,  
$$
where 
\begin{equation}
	I_1:=\int_{t_0}^{t_1}f(t)\,\frac{dt}t,\quad I_2:=\int_{t_1}^{t_2}f(t)\,\frac{dt}{1-t}, 
\end{equation}
\begin{equation}
	f(t):=t^{k+1}(1-t)^{n-k+1},\quad t_0:=\tfrac k{n+2}, \quad t_1:=\tfrac k{n+1}, \quad t_2:=\tfrac{k+1}{n+2}, 
\end{equation}
so that $0\le t_0<t_1<t_2<1$ and the function $f$ is positive and increasing on the interval $(t_0,t_2)$. 

It remains to check the inequality $I_1\le I_2$. The case $k=0$ is trivial. So, assume that $k>0$. Since each side of the inequality $I_1\le I_2$ is linear in $f$, it is enough to prove if for functions $f$ of the form $t\mapsto I\{t>w\}$, where $I$ is the indicator function and $w$ is any real number in the interval $[t_0,t_2]$. If $w\ge t_1$, then $I_1-I_2=0-I_2\le0$. If $w\in(t_0,t_1)$, then $I_1$ can be increased by moving $w$ to the left, without affecting $I_2$. So, without loss of generality $w=t_0$ and $f=1$, in which case $I_1-I_2=0$. $\qed$