Spectral radius inequality for normal matrices. I've read in several places now that $\rho(AB) \leq \rho(A)\rho(B)$ holds if the matrices are normal. I've scoured the web for a proof since I can't seem to get there myself. How would one go about to prove this statement? Links to proof is also welcome.
 A: Note that for any matrix, the induced 2-norm of that matrix is defined as $$\|A\|_2:=\sup_{x\neq0}\frac{\|Ax\|_2}{\|x\|_2},$$ and thus by the spectral theorem for normal matrices, there exists a basis of eigenvectors of $A$ for $\mathbb{C}^{n\times n}$, and thus $\|A\|_2=\rho(A)$. Similarly $\|B\|_2=\rho(B)$. For $AB$, which is not necessarily normal, we hace $\|AB\|_2=\sigma_{\mathrm{max}}(AB)$, the maximal singular value of $AB$, see also this post.
Now using this, we can see directly that $$\rho(AB)\leq\sigma_{\mathrm{max}}(AB)=\|AB\|_2\leq\|A\|_2\|B\|_2=\rho(A)\rho(B),$$ where the second inequality holds because the induced $2$-norm $\|\cdot\|$ is indeed a norm.
For the first inequality, observe that for any induced matrix $p$-norm, we have  for any $v\in\mathbb{C}^n$, and therefore also for $v$ the eigenvector with norm $1$ corresponding to $|\lambda|=\rho(A)$: $$\|A\|_p:=\sup_{x\neq0}\frac{\|Ax\|_p}{\|x\|_p}=\sup_{\|x\|_p=1}\|Ax\|_p\geq\|Av\|_p=\|\lambda v\|_p=\rho(A)\|v\|_p=\rho(A).$$
