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The answer by Piotr Migdal can be modified to give the accurate inequality $$r([A,B])\le4r(A)r(B),\qquad\forall A,B\in{\bf M}_n(\mathbb C).$$ The only new argument is that for every matrix $M$, there exists an angle $\theta$ such that $r(M)=\|{\rm Re}(e^{-i\theta}M)\|_2.$
Actually, we do have $$r(M)=\sup_\alpha\|{\rm Re}(e^{-i\alpha}M)\|_2.$$ Hereabove, the real part is defined as ${\rm Re} N=\frac12(N+\bar N^T)$. Notice that I employ the notation $\|\cdot\|$ (operator norm) which coincides with $\sigma_1$.
Let us apply this to $M=[A,B]$. With $\theta$ as above, let us decompose $e^{-i\theta}A=A_{\theta h}+iA_{\theta a}$. Then let us proceed as Piotr did: $$r([A,B])=\|{\rm Re}[e^{-i\theta}A,B]\| = \|[A_{\theta h},B_h]-[A_{\theta a},B_a]\|\le2(\|A_{\theta h}\|\cdot\|B_h\|+\|A_{\theta a}\|\cdot\|B_a\|),$$ |i[A_{\theta h},B_a]+i[A_{\theta a},B_h]\|\le2(\|A_{\theta h}\|\cdot\|B_a\|+\|A_{\theta a}\|\cdot\|B_h\|),$$which gives$$r([A,B])\le4r(e^{-i\theta}A)r(B).$$Hence the result. 1 The answer by Piotr Migdal can be modified to give the accurate inequality$$r([A,B])\le4r(A)r(B),\qquad\forall A,B\in{\bf M}_n(\mathbb C).$$The only new argument is that for every matrix M, there exists an angle \theta such that r(M)=\|{\rm Re}(e^{-i\theta}M)\|_2. Actually, we do have$$r(M)=\sup_\alpha\|{\rm Re}(e^{-i\alpha}M)\|_2.$$Hereabove, the real part is defined as {\rm Re} N=\frac12(N+\bar N^T). Notice that I employ the notation \|\cdot\| (operator norm) which coincides with \sigma_1. Let us apply this to M=[A,B]. With \theta as above, let us decompose e^{-i\theta}A=A_{\theta h}+iA_{\theta a}. Then let us proceed as Piotr did:$$r([A,B])=\|{\rm Re}[e^{-i\theta}A,B]\| = \|[A_{\theta h},B_h]-[A_{\theta a},B_a]\|\le2(\|A_{\theta h}\|\cdot\|B_h\|+\|A_{\theta a}\|\cdot\|B_a\|),$$which gives$$r([A,B])\le4r(e^{-i\theta}A)r(B). Hence the result.