TITLE: Bounding the commutator [A,B] in terms of the numerical radius
QUESTION [17 upvotes]: Given a norm $N$ over ${\bf M}_n(\mathbb C)$, it is a natural question to find the best constant $C_N$ such that
$$N([A,B])\le C_N N(A)N(B),\qquad\forall A,B\in{\bf M}_n(\mathbb C).$$
The answer is known at least in the following cases:

the operator norm $\|A\|_2=\sup\frac{\|Ax\|_2}{\|x\|_2}$ where the norm over $\mathbb C^n$ is the standard Hermitian $\|x\|_2^2=\sum_j|x_j|^2$. Then
$$\|[A,B]\|_2\le2\|A\|_2\|B\|_2$$
is optimal for $n\ge2$.
the Frobenius norm $\|A\|^2_F=\sum_{i,j}|a_{ij}|^2$. Then a theorem by Böttcher & Wentzel (2008) tells us that
$$\|[A,B]\|_F\le\sqrt2\|A\|_F\|B\|_F,$$
and again this is optimal.


I have a third norm in mind, yet of a different nature: the numerical radius
  $$r(A)=\sup_{x\ne0}\frac{|x^*Ax|}{\|x\|^2}.$$
  This is the smallest radius of a disk $D(0;r)$ containing the numerical range (or Hausdorffian) of the matrix.
  What is the optimal constant $C_{nr}$ such that $r([A,B])\le C_{nr}r(A)r(B)$ for all $A,B$ in ${\bf M}_n(\mathbb C)$ ?

Let me point out that $r$ is not submultiplicative. We have at best $r(MN)\le 4r(M)r(N)$, which gives by the triangle inequality $r([A,B])\le8r(A)r(B)$, but this is certainly not optimal. However, it is a super-stable norm, in the sense that $r(M^k)\le r(M)^k$ for every $k\ge1$.
This question naturally extends to $n$-commutators, in the spirit of my previous question Standard polynomials applied to matrices .
Edit. See below Piotr Migdal's answer and my adaptation of it. It gives $C_{nr}=4$.

REPLY [6 votes]: 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:
$$r([A,B])=\|{\rm Re}[e^{-i\theta}A,B]\| = \|i[A_{\theta h},B_a]+i[A_{\theta a},B_h]\|\le2(\|A_{\theta h}\|\cdot\|B_a\|+\|A_{\theta a}\|\cdot\|B_h\|),$$
which gives 
$$r([A,B])\le4r(e^{-i\theta}A)r(B).$$
Hence the result.