Let $B(p)$ denote the Bernoulli distribution over $\{0,1\}$ and $B(p)^n$ the corresponding product distribution over $\{0,1\}^n$. For $n>1$ and $0<x<1$, define
$$P\_n(x):=B(\frac12+\frac x2)^n$$
and
$$Q\_n(x):=B(\frac12-\frac x2)^n.$$
I can show $ ||P\_n(x)-Q\_n(x)||\_1 \le 2.1\sqrt{n}x$ using Pinsker's inequality.


**Conjecture:**
$$ ||P\_n(x)-Q\_n(x)||\_1 \le \sqrt{3n}x.$$



Sharpened Pinsker inequality for special case
Looking at **[[Pinsker's inequality](https://en.wikipedia.org/wiki/Pinsker's_inequality)]**, I am assuming that the left-hand side of the conjectured inequality is understood as the the total variation norm of the signed measure $P\_n(x) - Q\_n(x)$, which equals 
\begin{equation}
 s(y):=s\_n(y):=2\sum\_{j=0}^m\binom nj\big[\big(\tfrac12 + y\big)^{n - j} \big(\tfrac12 - y\big)^j - \big(\tfrac12 - y\big)^{n - j} \big(\tfrac12 + y\big)^j\big], 
\end{equation}
where 
\begin{equation}
m:=\lfloor n/2\rfloor,\quad y:=x/2\in[0,1/2). 
\end{equation}
Using identities $(n-j)\binom nj=n\binom{n-1}j$ and $j\binom nj=n\binom{n-1}{j-1}$, we have 
\begin{equation}
 \tfrac1{2n}\,s'(y):=\sum\_{j=0}^m\binom{n-1}j\big[\big(\tfrac12 + y\big)^{n - j-1} \big(\tfrac12 - y\big)^j + \big(\tfrac12 - y\big)^{n - j-1} \big(\tfrac12 + y\big)^j\big] 
\end{equation}
\begin{equation} 
 -\sum\_{j=1}^m\binom{n-1}{j-1}[\big(\tfrac12 + y\big)^{n - j} \big(\tfrac12 - y\big)^{j-1} + \big(\tfrac12 - y\big)^{n - j} \big(\tfrac12 + y\big)^{j-1}\big]. 
\end{equation}
Making in the second sum the substitution $j=k+1$ and then replacing there $k$ back by $j$, we have 
\begin{equation}
 \tfrac1{2n}\,s'(y)\Big/\binom{n-1}m=\big(\tfrac12 + y\big)^{n - m-1} \big(\tfrac12 - y\big)^m + \big(\tfrac12 - y\big)^{n - m-1} \big(\tfrac12 + y\big)^m
 =2^{2-n}(1-4y^2)^{n-m-1}, 
\end{equation}
which is non-increasing in $y\in[0,1/2)$ for $n=1,2,\dots$. 
So, $s$ is concave. 


Moreover, 
\begin{equation}
 s'\_n(0)=n2^{3-n}\binom{n-1}{\lfloor n/2\rfloor}\le 2A\sqrt n \tag{\*}
\end{equation}
for some universal real constant $A>0$. It should be easy to see that one can take ($A=2$ for $n=1$ and) $A=\sqrt3$ for $n=2,3,\dots$. 


Since $s(0)=0$ and $s$ is concave, it will then follow that for $n=2,3,\dots$
$$\|P\_n(x) - Q\_n(x)\|=s(y)\le2Ay\sqrt n=\sqrt{3n}\,x,$$
as desired (actually, for all $x\in[0,1]$). 


**Addendum:** To verify $(\*)$, let 
\begin{equation}
 r\_n:=s'\_n(0)/(2\sqrt n). 
\end{equation}
We need to show that $r\_n\le\sqrt3$ for $n=2,3,\dots$. For $m=1,2,\dots$, 
\begin{equation}
 \frac{r\_{2m+2}}{r\_{2m}}=\sqrt{\frac{4m^2+4m+1}{4m^2+4m}}>1,
\end{equation}
so that $r\_{2m}$ increases in $m$, and $r\_{2m}\to2\sqrt{2/\pi}$ as $m\to\infty$. 
So, $r\_{2m}<2\sqrt{2/\pi}<\sqrt3$ for all $m=1,2,\dots$. 
Similarly, in the "odd" case, for $m=2,3,\dots$ 
\begin{equation}
 \frac{r\_{2m+3}}{r\_{2m+1}}=\sqrt{1-\frac{6 m^3+11 m^2-8 m-4}{4(2m+1)(m^2-1)^2}}<1,
\end{equation}
so that $r\_{2m+1}$ decreases in $m\ge2$. It remains to note that $(r\_1,r\_3,r\_5)=(\sqrt4, \sqrt3, \sqrt{45/16})$, so that $r\_1>r\_3>r\_5$. 


