Set $A:=C\_0((0,1]) \* C\_0((0,1])$ (the free product C\*-algebra), with canonical generators $a,b$ (positive contractions). Does there exists some $\gamma>0$ such that, for any $x,y \in A$ if $x^\*x=a$ and $y^\*y=b$ then
$$ \|[xx^\*,yy^\*]\| > \gamma? $$



Are free positive operators equivalent to almost-commuting operators?
YES. (However, the answer for the same question for von Neumann algebras is NO.) 
I take $$A:=\lbrace f\in C([0,1],M\_2) : f(0), f(1) \in D\_2\rbrace.$$ 
Here $D\_2\cong\ell\_\infty^2$ is the diagonal. 
Let $$Q:=\mathrm{ev}\_0\oplus\mathrm{ev}\_1\colon A\to\ell\_\infty^2\oplus\ell\_\infty^2.$$
Let $$a=\left(\begin{matrix} 1 & 0\\ 0 & \frac{1}{2}\end{matrix}\right)\mbox{ (constant) and }
b=\left(\begin{matrix} t & \sqrt{t(1-t)}\\ \sqrt{t(1-t)} & 1-t\end{matrix}\right)\mbox{ (projection)}.$$
Suppose $a=x^\*x$ and $b=y^\*y$. Then, $x=u|x|$ and $u\in A$ (unitary). One has
$$\|[xx^\*,yy^\*]\|=\|[a,u^\*yy^\*u]\|.$$
Since $Q(A)$ is commutative, $u^\*yy^\*u$ is a {\bf projection} such that 
$Q(u^\*yy^\*u)=Q(b)=\mathrm{diag}(0,1)\oplus\mathrm{diag}(1,0)$.
This implies $\|[a,u^\*yy^\*u]\|\geq 1/4$. 


