Theorem:
Let $\X$ be a complex Euclidean space, let $\rho_0,\rho_1\in\Density(\X)$ be density operators satisfying
\[
\bignorm{\rho_0 - \rho_1}_1 \geq \varepsilon
\]
for $\varepsilon > 0$, and let $n$ be an arbitrary positive integer.
Then
\[
\Bignorm{\rho_0^{\otimes n} - \rho_1^{\otimes n}}_1
\geq 2 - 2 \exp\biggl(-\frac{n\varepsilon^2}{8}\biggr).
\]
(The notation $\rho^{\otimes n}$ means $\rho$ tensored with itself $n$ times. For example, $\rho^{\otimes 4} = \rho\otimes\rho\otimes\rho\otimes\rho$.)
Proof:
By the Fuchs-van de Graaf inequalities (Theorem 3.34) we have that the following two statements are equivalent:
\[
1-\frac{1}{2}\bignorm{\rho_0-\rho_1}_1\leq\fid(\rho_0,\rho_1)\leq\sqrt{1-\frac{1}{4}\bignorm{\rho_0-\rho_1}_1^2},
\]
\[
2-2\fid(\rho_0,\rho_1)\leq\bignorm{\rho_0-\rho_1}_1\leq 2\sqrt{1-\fid(\rho_0,\rho_1)^2}.
\]
Also, by Proposition 3.16, it follows that
\[
\fid(\rho_0^{\otimes n},\rho_1^{\otimes n})=\fid(\rho_0,\rho_1)^{n}.
\]
Therefore,
\[
\fid(\rho_0^{\otimes n},\rho_1^{\otimes n})^2=\fid(\rho_0,\rho_1)^{2n}\leq\left(1-\frac{1}{4}\bignorm{\rho_0-\rho_1}_1^2\right)^n\leq\left(1-\frac{1}{4}\varepsilon^2\right)^n,
\]
since $ \varepsilon\leq\bignorm{\rho_0 - \rho_1}_1$ by assumption.
Since for $x\in\mathbb{R}$ such that $0\leq x\leq1$, and any positive integer $n$,
\[
(1-x)^n\leq \exp(-nx),
\]
then for $0\leq\varepsilon\leq2$,
\[
\left(1-\frac{1}{4}\varepsilon^2\right)^n\leq \exp\biggl(-\frac{n\varepsilon^2}{4}\biggr).
\]
This implies that
\[
\fid(\rho_0^{\otimes n},\rho_1^{\otimes n})\leq\left(1-\frac{1}{4}\varepsilon^2\right)^{n/2}\leq\exp\biggl(-\frac{n\varepsilon^2}{8}\biggr).
\]
However, by the Fuchs-van de Graaf inequalities, since
\[
2-2\fid(\rho_0^{\otimes n},\rho_1^{\otimes n})\leq\bignorm{\rho_0^{\otimes n}-\rho_1^{\otimes n}}_1,
\]
the previous result implies
\[
2-2\exp\biggl(-\frac{n\varepsilon^2}{8}\biggr)\leq2-2\fid(\rho_0^{\otimes n},\rho_1^{\otimes n})\leq\bignorm{\rho_0^{\otimes n}-\rho_1^{\otimes n}}_1,
\]
which completes the proof.
Let $\X$ be a complex Euclidean space, let $\rho_0,\rho_1\in\Density(\X)$ be density operators satisfying
\[
\bignorm{\rho_0 - \rho_1}_1 \geq \varepsilon
\]
for $\varepsilon > 0$, and let $n$ be an arbitrary positive integer.
Then
\[
\Bignorm{\rho_0^{\otimes n} - \rho_1^{\otimes n}}_1
\geq 2 - 2 \exp\biggl(-\frac{n\varepsilon^2}{8}\biggr).
\]
(The notation $\rho^{\otimes n}$ means $\rho$ tensored with itself $n$ times. For example, $\rho^{\otimes 4} = \rho\otimes\rho\otimes\rho\otimes\rho$.)
Proof:
By the Fuchs-van de Graaf inequalities (Theorem 3.34) we have that the following two statements are equivalent:
\[
1-\frac{1}{2}\bignorm{\rho_0-\rho_1}_1\leq\fid(\rho_0,\rho_1)\leq\sqrt{1-\frac{1}{4}\bignorm{\rho_0-\rho_1}_1^2},
\]
\[
2-2\fid(\rho_0,\rho_1)\leq\bignorm{\rho_0-\rho_1}_1\leq 2\sqrt{1-\fid(\rho_0,\rho_1)^2}.
\]
Also, by Proposition 3.16, it follows that
\[
\fid(\rho_0^{\otimes n},\rho_1^{\otimes n})=\fid(\rho_0,\rho_1)^{n}.
\]
Therefore,
\[
\fid(\rho_0^{\otimes n},\rho_1^{\otimes n})^2=\fid(\rho_0,\rho_1)^{2n}\leq\left(1-\frac{1}{4}\bignorm{\rho_0-\rho_1}_1^2\right)^n\leq\left(1-\frac{1}{4}\varepsilon^2\right)^n,
\]
since $ \varepsilon\leq\bignorm{\rho_0 - \rho_1}_1$ by assumption.
Since for $x\in\mathbb{R}$ such that $0\leq x\leq1$, and any positive integer $n$,
\[
(1-x)^n\leq \exp(-nx),
\]
then for $0\leq\varepsilon\leq2$,
\[
\left(1-\frac{1}{4}\varepsilon^2\right)^n\leq \exp\biggl(-\frac{n\varepsilon^2}{4}\biggr).
\]
This implies that
\[
\fid(\rho_0^{\otimes n},\rho_1^{\otimes n})\leq\left(1-\frac{1}{4}\varepsilon^2\right)^{n/2}\leq\exp\biggl(-\frac{n\varepsilon^2}{8}\biggr).
\]
However, by the Fuchs-van de Graaf inequalities, since
\[
2-2\fid(\rho_0^{\otimes n},\rho_1^{\otimes n})\leq\bignorm{\rho_0^{\otimes n}-\rho_1^{\otimes n}}_1,
\]
the previous result implies
\[
2-2\exp\biggl(-\frac{n\varepsilon^2}{8}\biggr)\leq2-2\fid(\rho_0^{\otimes n},\rho_1^{\otimes n})\leq\bignorm{\rho_0^{\otimes n}-\rho_1^{\otimes n}}_1,
\]
which completes the proof.