[proofplan]
We choose the common dominating measure $\mu := P+Q$, so the densities $p := dP/d\mu$ and $q := dQ/d\mu$ satisfy $p+q=1$ $\mu$-almost everywhere. With this normalization, total variation has the exact representation $1-\operatorname{TV}(P,Q)=\int_\Omega \min\{p,q\}\,d\mu$. The Hellinger affinity is bounded by Cauchy's inequality because $\sqrt{pq}\leq \sqrt{\min\{p,q\}}$ and $\mu(\Omega)=2$. Squaring gives $\rho_H(P,Q)^2\leq 2(1-\operatorname{TV}(P,Q))$, which is the desired inequality.
[/proofplan]
[step:Choose the joint dominating measure and record the density identities]
Define the finite measure $\mu$ on $(\Omega,\mathcal{F})$ by
\begin{align*}
\mu(A) := P(A)+Q(A), \qquad A \in \mathcal{F}.
\end{align*}
Then $P \ll \mu$ and $Q \ll \mu$. Let $p:\Omega \to [0,\infty)$ be the Radon--Nikodym density $p:=dP/d\mu$, and let $q:\Omega \to [0,\infty)$ be the Radon--Nikodym density $q:=dQ/d\mu$. Since $\mu=P+Q$, the measure with density $p+q$ with respect to $\mu$ equals $\mu$ itself. Hence
\begin{align*}
p+q=1
\end{align*}
$\mu$-almost everywhere. Also
\begin{align*}
\mu(\Omega)=P(\Omega)+Q(\Omega)=2.
\end{align*}
[/step]
[step:Rewrite the complement of total variation as the overlap integral]
For every $A\in\mathcal{F}$,
\begin{align*}
P(A)-Q(A)=\int_A (p-q)\,d\mu.
\end{align*}
Let $E:=\{\omega\in\Omega:p(\omega)>q(\omega)\}$. Since $p$ and $q$ are measurable, $E\in\mathcal{F}$. For each $B\in\mathcal{F}$, let $\mathbb{1}_B:\Omega\to\{0,1\}$ denote the indicator function of $B$, defined by $\mathbb{1}_B(\omega)=1$ if $\omega\in B$ and $\mathbb{1}_B(\omega)=0$ if $\omega\notin B$. Define the positive and negative parts of the measurable function $p-q:\Omega\to\mathbb{R}$ by
\begin{align*}
(p-q)^+(\omega):=\max\{p(\omega)-q(\omega),0\}, \qquad (p-q)^-(\omega):=\max\{q(\omega)-p(\omega),0\}.
\end{align*}
For every $A\in\mathcal{F}$, the pointwise inequalities $(p-q)\mathbb{1}_A\leq (p-q)^+$ and $(p-q)\mathbb{1}_E=(p-q)^+$ give
\begin{align*}
P(A)-Q(A)\leq \int_\Omega (p-q)^+\,d\mu
\end{align*}
and
\begin{align*}
P(E)-Q(E)=\int_\Omega (p-q)^+\,d\mu.
\end{align*}
It remains to control the opposite sign in the absolute value. Since $P(\Omega)=Q(\Omega)=1$, we have
\begin{align*}
\int_\Omega (p-q)\,d\mu=P(\Omega)-Q(\Omega)=0.
\end{align*}
Because $(p-q)=(p-q)^+-(p-q)^-$ and both positive and negative parts are integrable, this gives
\begin{align*}
\int_\Omega (p-q)^+\,d\mu=\int_\Omega (p-q)^-\,d\mu.
\end{align*}
For every $A\in\mathcal{F}$,
\begin{align*}
Q(A)-P(A)=\int_A (q-p)\,d\mu\leq \int_\Omega (q-p)^+\,d\mu=\int_\Omega (p-q)^-\,d\mu=\int_\Omega (p-q)^+\,d\mu.
\end{align*}
Therefore $|P(A)-Q(A)|\leq \int_\Omega (p-q)^+\,d\mu$ for every $A\in\mathcal{F}$, and equality is attained at $E$. Thus $\operatorname{TV}(P,Q)=\int_\Omega (p-q)^+\,d\mu$. The positive part of $p-q$ is $(p-q)^+=p-\min\{p,q\}$, so
\begin{align*}
\operatorname{TV}(P,Q)=\int_\Omega \bigl(p-\min\{p,q\}\bigr)\,d\mu.
\end{align*}
Since $\int_\Omega p\,d\mu=P(\Omega)=1$, this becomes
\begin{align*}
\operatorname{TV}(P,Q)=1-\int_\Omega \min\{p,q\}\,d\mu.
\end{align*}
Therefore
\begin{align*}
1-\operatorname{TV}(P,Q)=\int_\Omega \min\{p,q\}\,d\mu.
\end{align*}
[guided]
We want to express $1-\operatorname{TV}(P,Q)$ as a quantity that can be compared directly with the Hellinger affinity. The natural object is the common mass of the two densities, namely the integral of their pointwise minimum.
For each measurable set $A\in\mathcal{F}$, the Radon--Nikodym representations give
\begin{align*}
P(A)-Q(A)=\int_A (p-q)\,d\mu.
\end{align*}
To identify the supremum in the definition of total variation, define $E:=\{\omega\in\Omega:p(\omega)>q(\omega)\}$. This set is measurable because $p$ and $q$ are measurable. For each $B\in\mathcal{F}$, let $\mathbb{1}_B:\Omega\to\{0,1\}$ denote the indicator function of $B$, defined by $\mathbb{1}_B(\omega)=1$ if $\omega\in B$ and $\mathbb{1}_B(\omega)=0$ if $\omega\notin B$. Define the positive and negative parts of the measurable function $p-q:\Omega\to\mathbb{R}$ by
\begin{align*}
(p-q)^+(\omega):=\max\{p(\omega)-q(\omega),0\}, \qquad (p-q)^-(\omega):=\max\{q(\omega)-p(\omega),0\}.
\end{align*}
These functions separate the signed density difference into its upward and downward deviations. For every $A\in\mathcal{F}$, the pointwise inequality $(p-q)\mathbb{1}_A\leq (p-q)^+$ gives
\begin{align*}
P(A)-Q(A)\leq \int_\Omega (p-q)^+\,d\mu.
\end{align*}
On the other hand, $(p-q)\mathbb{1}_E=(p-q)^+$, so
\begin{align*}
P(E)-Q(E)=\int_\Omega (p-q)^+\,d\mu.
\end{align*}
This proves the largest positive deviation, but total variation uses an absolute value, so we must also rule out a larger negative deviation. Since both $P$ and $Q$ are probability measures,
\begin{align*}
\int_\Omega (p-q)\,d\mu=P(\Omega)-Q(\Omega)=0.
\end{align*}
Writing $(p-q)=(p-q)^+-(p-q)^-$ and using integrability of the positive and negative parts, we obtain
\begin{align*}
\int_\Omega (p-q)^+\,d\mu=\int_\Omega (p-q)^-\,d\mu.
\end{align*}
Now for every $A\in\mathcal{F}$,
\begin{align*}
Q(A)-P(A)=\int_A(q-p)\,d\mu\leq \int_\Omega(q-p)^+\,d\mu=\int_\Omega(p-q)^-\,d\mu=\int_\Omega(p-q)^+\,d\mu.
\end{align*}
Thus both $P(A)-Q(A)$ and $Q(A)-P(A)$ are bounded above by the same quantity, so
\begin{align*}
|P(A)-Q(A)|\leq \int_\Omega (p-q)^+\,d\mu
\end{align*}
for every $A\in\mathcal{F}$. Since equality is attained at $E$,
\begin{align*}
\operatorname{TV}(P,Q)=\int_\Omega (p-q)^+\,d\mu.
\end{align*}
Pointwise, the positive part satisfies
\begin{align*}
(p-q)^+=p-\min\{p,q\}.
\end{align*}
Indeed, if $p\geq q$, then $(p-q)^+=p-q=p-\min\{p,q\}$; if $p<q$, then both sides are $0$. Therefore
\begin{align*}
\operatorname{TV}(P,Q)=\int_\Omega \bigl(p-\min\{p,q\}\bigr)\,d\mu.
\end{align*}
By linearity of the integral for non-negative finite integrable functions,
\begin{align*}
\operatorname{TV}(P,Q)=\int_\Omega p\,d\mu-\int_\Omega \min\{p,q\}\,d\mu.
\end{align*}
Since $p=dP/d\mu$ and $P$ is a probability measure,
\begin{align*}
\int_\Omega p\,d\mu=P(\Omega)=1.
\end{align*}
Substituting this gives
\begin{align*}
\operatorname{TV}(P,Q)=1-\int_\Omega \min\{p,q\}\,d\mu,
\end{align*}
or equivalently
\begin{align*}
1-\operatorname{TV}(P,Q)=\int_\Omega \min\{p,q\}\,d\mu.
\end{align*}
This identity converts the problem into proving that the overlap integral dominates one half of the square of the Hellinger affinity.
[/guided]
[/step]
[step:Bound the Hellinger affinity by the overlap integral]
Define the measurable function $m:\Omega \to [0,\infty)$ by $m(\omega):=\min\{p(\omega),q(\omega)\}$.
Since $p+q=1$ $\mu$-almost everywhere, we have $0\leq p\leq 1$ and $0\leq q\leq 1$ $\mu$-almost everywhere. Hence
\begin{align*}
pq \leq m
\end{align*}
$\mu$-almost everywhere, because if $p\leq q$ then $pq\leq p=m$, and if $q\leq p$ then $pq\leq q=m$. We justify that the Hellinger affinity may be computed with this particular dominating measure. Let $\nu$ be any common dominating measure for $P$ and $Q$, and let $r:\Omega\to[0,\infty)$ and $s:\Omega\to[0,\infty)$ denote the Radon--Nikodym densities $r:=dP/d\nu$ and $s:=dQ/d\nu$. Define the finite measure $\theta:=\mu+\nu$, and let $u:\Omega\to[0,\infty)$ and $v:\Omega\to[0,\infty)$ denote the Radon--Nikodym densities $u:=d\mu/d\theta$ and $v:=d\nu/d\theta$. By the chain rule for Radon--Nikodym derivatives, the $\theta$-densities of $P$ and $Q$ are both given by
\begin{align*}
p u=r v, \qquad q u=s v
\end{align*}
$\theta$-almost everywhere. Hence
\begin{align*}
\sqrt{pq}\,u=\sqrt{rs}\,v
\end{align*}
$\theta$-almost everywhere, with both sides equal to the common function $\sqrt{(pu)(qu)}=\sqrt{(rv)(sv)}$. Using the change-of-density formula for non-negative [measurable functions](/page/Measurable%20Functions),
\begin{align*}
\int_\Omega \sqrt{pq}\,d\mu=\int_\Omega \sqrt{pq}\,u\,d\theta=\int_\Omega \sqrt{rs}\,v\,d\theta=\int_\Omega \sqrt{rs}\,d\nu.
\end{align*}
Thus the value of the affinity is independent of the common dominating measure. Taking $\nu$ to be the measure used in the definition of $\rho_H(P,Q)$ gives
\begin{align*}
\rho_H(P,Q)=\int_\Omega \sqrt{pq}\,d\mu.
\end{align*}
Using $pq\leq m$ and monotonicity of the [Lebesgue integral](/page/Lebesgue%20Integral) for non-negative measurable functions,
\begin{align*}
\rho_H(P,Q)\leq \int_\Omega \sqrt{m}\,d\mu.
\end{align*}
Moreover $m\leq p$, so $\int_\Omega m\,d\mu\leq \int_\Omega p\,d\mu=1$; hence $\sqrt{m}\in L^2(\Omega,\mathcal{F},\mu)$. Since $\mu(\Omega)=2$, the constant function $1:\Omega\to\mathbb{R}$ belongs to $L^2(\Omega,\mathcal{F},\mu)$. To obtain the needed product estimate without invoking an external result, consider the non-negative quadratic polynomial
\begin{align*}
\int_\Omega \bigl(\sqrt{m}-\lambda\bigr)^2\,d\mu
\end{align*}
in the real parameter $\lambda\in\mathbb{R}$. Expanding the square with respect to the measure $\mu$ gives
\begin{align*}
0\leq \lambda^2\int_\Omega 1\,d\mu-2\lambda\int_\Omega \sqrt{m}\,d\mu+\int_\Omega m\,d\mu.
\end{align*}
A real quadratic polynomial that is non-negative for every $\lambda\in\mathbb{R}$ has non-positive discriminant, so
\begin{align*}
\left(\int_\Omega \sqrt{m}\,d\mu\right)^2\leq \left(\int_\Omega m\,d\mu\right)\left(\int_\Omega 1\,d\mu\right).
\end{align*}
Since $\int_\Omega 1\,d\mu=\mu(\Omega)=2$, we obtain
\begin{align*}
\left(\int_\Omega \sqrt{m}\,d\mu\right)^2\leq 2\int_\Omega m\,d\mu.
\end{align*}
Thus
\begin{align*}
\rho_H(P,Q)^2 \leq 2\int_\Omega \min\{p,q\}\,d\mu.
\end{align*}
[/step]
[step:Substitute the total variation identity to finish]
Using the identity from the overlap step,
\begin{align*}
\rho_H(P,Q)^2\leq 2\int_\Omega \min\{p,q\}\,d\mu.
\end{align*}
Substituting $\int_\Omega \min\{p,q\}\,d\mu=1-\operatorname{TV}(P,Q)$ gives
\begin{align*}
\rho_H(P,Q)^2\leq 2\bigl(1-\operatorname{TV}(P,Q)\bigr).
\end{align*}
Dividing by $2$ gives
\begin{align*}
1-\operatorname{TV}(P,Q)\geq \frac{1}{2}\rho_H(P,Q)^2.
\end{align*}
This is the claimed bound.
[/step]
