**Yes.** We have a necessary and sufficient characterization for when the Cauchy-Schwarz inequality becomes equality.

For this post, $U,V,W$ shall denote finite dimensional complex Hilbert spaces. Suppose that $A_1,\dots,A_n:U\rightarrow U,B_1,\dots,B_n:V\rightarrow V$ are linear operators. Then define an operator $\Gamma(A_1,\dots,A_n;B_1,\dots,B_n):L(V,U)\rightarrow L(V,U)$ by letting $$\Gamma(A_1,\dots,A_n;B_1,\dots,B_n)(X)=A_1XB_1^*+\dots+A_nXB_n^*.$$

Observe that $\Gamma(A_1,\dots,A_n;B_1,\dots,B_n)$ is similar to $A_1\otimes \overline{B_1}+\dots+A_n\otimes\overline{B_n}$ (Here, $\overline{C}=(C^{T})^{*}=(C^*)^T$, so $\overline{C}$ is the matrix obtained by replacing every entry in $C$ with its complex conjugate), so the spectral radius Cauchy-Schwarz inequality
$$\rho(\Gamma(A_1,\dots,A_n;B_1,\dots,B_n))
\leq\rho(\Phi(A_1,\dots,A_n))^{1/2}\cdot\rho(\Phi(B_1,\dots,B_n))^{1/2}$$
hold in all cases. It is not too hard to prove the spectral radius Cauchy-Schwarz inequality by using the conventional Cauchy-Schwarz inequality and the characterization of $\rho(\Phi(A_1,\dots,A_n))^{1/2}$ given in the 1998 paper The $p$-norm joint spectral radius for even integers by Ding-Xuan Zhou.

Observe that if there is a $\lambda$ and an invertible $B$ with $B_j=\lambda BA_jB^{-1}$ for $1\leq j\leq n$, then
$$\rho(\Gamma(A_1,\dots,A_n;B_1,\dots,B_n))
=\rho(\Phi(A_1,\dots,A_n))^{1/2}\cdot\rho(\Phi(B_1,\dots,B_n))^{1/2}.$$

I claim that when $(A_1,\dots,A_n)$ and $(B_1,\dots,B_n)$ have no invariant subspaces, this is the only way in which the spectral radius Cauchy-Schwarz inequality becomes equality. By decomposing $(A_1,\dots,A_n)$ and $(B_1,\dots,B_n)$ according to their invariant subspaces, we obtain necessary and sufficient conditions for when the Cauchy-Schwarz inequality is actually an equality.

Recall that a linear operator $\mathcal{E}:L(U)\rightarrow L(V)$ is called positive if $\mathcal{E}(P)$ is positive semidefinite whenever $P$ is positive semidefinite. A linear operator $\mathcal{E}:L(U)\rightarrow L(V)$ is called completely positive if
$\mathcal{E}\otimes 1_W:L(U\otimes W)\rightarrow L(V\otimes W)$ is positive whenever $W$ is a finite dimensional Hilbert space. We say that an operator $\mathcal{E}:L(U)\rightarrow L(V)$ is trace preserving if $\text{Tr}(\mathcal{E}(X))=\text{Tr}(X)$ for all $X\in L(U)$. An operator $\mathcal{E}:L(U)\rightarrow L(V)$ is said to be a quantum channel if $\mathcal{E}$ is both trace preserving and completely positive.

The completely positive mappings from $L(V)$ to $L(V)$ are precisely the mappings of the form $\Phi(A_1,\dots,A_n)$. Observe that any linear mapping from $L(V,U)$ to $L(V,U)$ is of the form $\Gamma(A_1,\dots,A_n;B_1,\dots,B_n)$. It is not too hard to show using the Frobenius inner product that the mapping $\Gamma(A_1,\dots,A_n;B_1,\dots,B_n)$ is trace preserving if and only if $A_1^*B_1+\dots+A_n^*B_n=1_V$. In particular, the quantum channels are precisely the mappings of the form $\Phi(A_1,\dots,A_n)$ where $A_1^*A_1+\dots+A_n^*A_n=1_V$.

Observe that if $\mathcal{E}$ is a quantum channel, then $\rho(\mathcal{E})=1$.

Let $E_{U,n}$ be the collection of tuples $(A_1,\dots,A_n)\in L(U)^n$ such that there is some complex number $\lambda$ and invertible $B$ such that if $B_j=\lambda BA_jB^{-1}$ for $1\leq j\leq n$, then $\Phi(B_1,\dots,B_n)$ is a quantum channel. By this answer, $L(U)^n\setminus E_{U,n}$ is a quite small set whenever $n>1$. In particular, if $(A_1,\dots,A_n)$ has no-invariant subspace, then $(A_1,\dots,A_n)\in E_{U,n}$.

Theorem: Suppose that $(A_1,\dots,A_n)\in L(U)^n,(B_1,\dots,B_n)\in L(V)^n$ have no invariant subspace. Then $$\rho(\Gamma(A_1,\dots,A_n;B_1,\dots,B_n))=\rho(\Phi(A_1,\dots,A_n))^{1/2}\cdot\rho(\Phi(B_1,\dots,B_n))^{1/2}$$ if and only if there is some invertible matrix $C$ along with some complex number $\eta\neq 0$ where
$A_j=\eta CB_jC^{-1}$ for $1\leq j\leq n$.

Proof: Since $(A_1,\dots,A_n),(B_1,\dots,B_n)$ have no invariant subspace, there are non-zero complex numbers $\mu,\nu$ along with invertible matrices $A,B$ where if we set $R_j=\mu AA_jA^{-1},S_j=\nu BB_jB^{-1}$ for $1\leq j\leq n$, then $\mathcal{E}=\Phi(R_1,\dots,R_n),\mathcal{F}=\Phi(S_1,\dots,S_n)$ are quantum channels. Let $W$ be a complex inner product space with orthonormal basis $(e_1,\dots,e_n).$ Now, let $R,S\in L(V,V\otimes W)$ be the operators defined by letting
$R=\sum_{j=1}^{n}R_j\otimes e_j,S=\sum_{j=1}^{n}S_j\otimes e_j$. Then
$$\mathcal{E}(X)=\text{Tr}_{W}(RXR^*),\mathcal{F}(X)=\text{Tr}_{W}(SXS^*).$$

Define a mapping $\mathcal{G}$ by setting $$\mathcal{G}(X)=\text{Tr}_{W}(RXS^*)=\sum_{k=1}^nR_kXS_k^*.$$

Since $R_1^*R_1+\dots+R_n^*R_n=S_1^*S_1+\dots+S_n^*S_n=1_V$, the mappings $R,S$ are isometries.

Now, let $\lambda$ be an eigenvalue of $\mathcal{G}$. Then there is some
eigenvector $X$ with $\mathcal{G}(X)=\lambda X$. Now perform a polar decomposition of $X$ to write $X=HP$ where $H$ is an isometry and $P$ is a positive semidefinite matrix. Therefore, we have
$$\lambda HP=\lambda X=\mathcal{G}(X)=\text{Tr}_{W}(RXS^*)=\text{Tr}_W(RHPS^*).$$ Now, set $T=(H^*\otimes 1_W)RH$. Then
$$\lambda P=\lambda H^*HP=H^*\text{Tr}_W(RHPS^*)=\text{Tr}_W((H^*\otimes 1_W)RHPS^*)=\text{Tr}_W(TPS^*).$$ Now let $P=\sum_{k=1}^{n}\sigma_ke_ke_k^*$. Then $\text{Tr}(\lambda P)=\lambda\cdot\sum_{k=1}^n\sigma_k$. Therefore, we have
$$\text{Tr}(\lambda P)=\text{Tr}(\text{Tr}_W(TPS^*))=\text{Tr}(TPS^*)=\sum_{k=1}^{n}\sigma_k\text{Tr}(Te_ke_k^*S^*)=\sum_{k=1}^n\sigma_k\text{Tr}(Te_k(Se_k)^*)=\sum_{k=1}^n\sigma_k\langle Te_k,Se_k\rangle.$$

Therefore, since $$\lambda\cdot\sum_{k=1}^n\sigma_k=\sum_{k=1}^n\sigma_k\langle Te_k,Se_k\rangle,$$ we know that $|\lambda|\leq 1$. Furthermore, if $|\lambda|=1$, then we know that $Te_k=\lambda Se_k$ whenever $\sigma_k>0$.
Therefore, we have $T|_{\text{Im}(P)}=\lambda\cdot S|_{\text{Im}(P)}$. In this case, we have $\lambda P=\text{Tr}_W(TPS^*)=\text{Tr}_W(\lambda SPS^*)$, so $P=\text{Tr}_W(SPS^*)$. Since
$P=\text{Tr}_W(SPS^*)$ and since $(S_1,\dots,S_n)$ has no invariant subspace, we know that $P$ is positive semidefinite. Therefore, $(H^*\otimes 1_W)RH=T=\lambda S$.

Now, if $v\in V$, then $\|RH v|=\|v\|=\|\lambda Sv\|=\|(H^*\otimes 1_W)RHv\|$. Therefore, if $v\in V$, then $RHv\in\text{Im}(H)\otimes 1_W$. Therefore, since $\text{Im}(H)$ is a non-trivial invariance subspace of $U$, we know that $\text{Im}(H)=U$. Therefore, $H$ is a unitary operator.

Thus, $H^*R_jH=\lambda S_j$ for $1\leq j\leq n$. Thus, $H^*\mu AA_jA^{-1}H=\lambda \nu BB_jB^{-1}$. We conclude that
$$A_j=\mu^{-1}\lambda\nu A^{-1}HBB_jB^{-1}H^{-1}A=\mu^{-1}\lambda\nu A^{-1}HBB_j(A^{-1}HB)^{-1}.$$

Q.E.D.

Theorem: Suppose that $A_1,\dots,A_n:U\rightarrow U,B_1,\dots,B_n:V\rightarrow V$ be linear operators. Then assign $U,V$ bases so that
$$A_j=\begin{bmatrix}
A_{j,1,1}&\dots&A_{j,1,u}\\
\vdots&\ddots&\vdots\\
A_{j,u,1}&\dots&A_{j,u,u}
\end{bmatrix}$$
and
$$B_j=\begin{bmatrix}
B_{j,1,1}&\dots&B_{j,1,v}\\
\vdots&\ddots&\vdots\\
B_{j,v,1}&\dots&B_{j,v,v}
\end{bmatrix}$$
and where for each $\alpha,\beta$ each of the $n$ matrices $A_{1,\alpha,\beta},\dots,A_{n,\alpha,\beta}$ have the same dimensions,
for each $\alpha,\beta$, each of the $n$ matrices $B_{1,\alpha,\beta},\dots,B_{n,\alpha,\beta}$ have the same dimensions, and where $A_{j,\alpha,\beta}=0$ whenever $\alpha>\beta$, and where $B_{j,\alpha,\beta}=0$ whenever $\alpha>\beta$, and where for $1\leq\alpha\leq u$, the matrices
$(A_{1,\alpha,\alpha},\dots,A_{n,\alpha,\alpha})$ have no non-trivial invariant subspace, and where if $1\leq\beta\leq v$, the matrices
$(B_{1,\beta,\beta},\dots,B_{n,\beta,\beta})$ have non-trivial no invariant subspace either.
Then $\rho(\Gamma(A_1,\dots,A_n;B_1,\dots,B_n))=\rho(\Phi(A_1,\dots,A_n))^{1/2}\cdot\rho(\Phi(B_1,\dots,B_n))^{1/2}$ if and only if there are $\alpha,\beta$ with $1\leq\alpha\leq u,1\leq\beta\leq v$ and where

$\rho(\Phi(A_1,\dots,A_n))=\rho(\Phi(A_{1,\alpha,\alpha},\dots,A_{n,\alpha,\alpha}))$

$\rho(\Phi(B_1,\dots,B_n))=\rho(\Phi(A_{1,\beta,\beta},\dots,A_{n,\beta,\beta}))$, and

There is an invertible matrix $C$ and some complex number $\lambda\neq 0$ such that $A_{j,\alpha,\beta}=\lambda CB_{j,\alpha,\beta}C^{-1}$ for $1\leq j\leq n$.

**A few observations:**

In this answer, we actually have a couple of different proofs of the spectral radius Cauchy-Schwarz inequality. To prove the Cauchy-Schwarz inequality, we need to prove that $\rho(\Gamma(R_1,\dots,R_n;S_1,\dots,S_n))\leq 1$ whenever $\Phi(R_1,\dots,R_n),\Phi(S_1,\dots,S_n)$ are quantum channels, and the general case will follow from the fact that $E_{V,n}$ is dense in $L(V)^n$ whenever $n>1$. We have already shown that $\rho(\Gamma(R_1,\dots,R_n;S_1,\dots,S_n))\leq 1$, but there is another way of showing this using the induced trace norm.

If $\mathcal{H}:L(V)\rightarrow L(V)$, then define the induced trace norm of $\mathcal{H}$ to be $\|\mathcal{H}\|_1=\max\{\|\mathcal{H}(X)\|_1:\|X\|_1\leq 1\}.$ Recall that if $R,S$ are isometries and $A$ is a complex matrix, then $\|RAS^*\|_1=\|A\|_1$ whenever the matrix multiplication exists. Also, recall that $\|\text{Tr}_{V}(X)\|_1\leq\|X\|_1$ whenever this inequality makes sense.

We have
$$\|\Gamma(R_1,\dots,R_n;S_1,\dots,S_n)(X)\|_1=\|\text{Tr}_{W}(RXS^*)\|_1\leq\|RXS^*\|_1=\|X\|_1.$$
Therefore, $\|\Gamma(R_1,\dots,R_n;S_1,\dots,S_n)\|_1\leq 1$, so
$\rho(\Gamma(R_1,\dots,R_n;S_1,\dots,S_n))\leq 1$ since the induced trace norm is submultiplicative.

**Empirical verification**

I have empirically verified using computer calculations that the conclusions that we have made are reasonable. Define a fitness function $F:M_d(\mathbb{C})^n\times M_d(\mathbb{C})^n\rightarrow\mathbb{R}$ by letting
$$F(A_1,\dots,A_n;B_1,\dots,B_n)=\frac{\rho(\Gamma(A_1,\dots,A_n;B_1,\dots,B_n))}{\rho(\Phi(A_1,\dots,A_n))^{1/2}\rho(\Phi(B_1,\dots,B_n))^{1/2}}.$$

One can maximize the value of $F(A_1,\dots,A_n;B_1,\dots,B_n)$ using gradient ascent to obtain examples of tuples $(A_1,\dots,A_n;B_1,\dots,B_n)$ with
$F(A_1,\dots,A_n;B_1,\dots,B_n)\approx 1$, but in each of these examples, we always have a complex number $\lambda$ along with some invertible $B$ where
$B_j\approx \lambda BA_jB^{-1}$ for $1\leq j\leq n$.