Let 
$$D\_{tv}(P,Q) = \frac{1}{2}\sum\_{a \in A}|P(a)-Q(a)|$$
and
$$D\_{\infty}(P,Q) = \sup\_{a \in A} \log \max\{\frac{P(a)}{Q(a)}, \frac{Q(a)}{P(a)}\},$$
where $P$ and $Q$ denote probability measures on a finite set $A$. 


Can you show that
$$D\_{tv}(P,Q) \leq 1- \exp(-D\_{\infty}(P,Q)) \leq D\_{\infty}(P,Q)?$$



Total variation and relative $\ell_\infty$ metric
The inequality $1- \exp(-D\_{\infty}(P,Q)) \leq D\_{\infty}(P,Q)$ is just an instance of the elementary inequality $e^u\ge1+u$ for all real $u$. 


Let us show an improved version of the inequality $D\_{tv}(P,Q) \leq 1- \exp(-D\_{\infty}(P,Q))$, namely,
\begin{equation\*}
 D\_{tv}(P,Q) \leq 1- \exp(-D\_{1,\infty}(P,Q)), 
\end{equation\*}
where 
\begin{equation\*}
 D\_{1,\infty}(P,Q) := \sup\_{a \in A} \ln\frac{P(a)}{Q(a)}
 \le D\_{\infty}(P,Q). 
\end{equation\*}


Indeed, for brevity let $r:=\exp D\_{1,\infty}(P,Q)[\ge1]$. Then $Q(a)\ge \frac1r\,P(a)$ for all $a\in A$. So, substituting $x=P(a)-Q(a)$ into the identity $x\_+:=\max(0,x)=\frac12\,(|x|+x)$ for all real $x$ and using the identity $\sum\_{a \in A}(P(a)-Q(a))=0$, we have 
\begin{multline\*}
 D\_{tv}(P,Q) = \sum\_{a \in A}(P(a)-Q(a))\_+
 \le\sum\_{a \in A}(1-\tfrac1r)P(a)=1-\tfrac1r \\ 
 =1- \exp(-D\_{1,\infty}(P,Q)), 
\end{multline\*}
as desired. 


