A version of von Neumann inequality Assume that $X,Y,Z$ are three commuting operators acting in a Hilbert space $H$. Assume also that they satisfy following properties:
1) $\|Z\| \le 1$, i.e. $Z$ is a contraction; 
2) For any complex scalar $z$ with $|z|\le 1$ we have $\|X+zY\| \le 1$. 
Does this imply that $\|X+ZY\| \le 1$? 
 A: I think the answer is "no," and a counterexample can be constructed as follows: we begin with a version of the Crabb-Davie counterexample to the 3-variable von Neumannn inequality. This consists of the polynomial
$$
p(z_1, z_2. z_3) = z_1^3+z_2^3+z_3^3 -z_1z_2z_3
$$
and three commuting $8\times 8$ matrices $T_1, T_2, T_3$ so that $\|p(T_1, T_2, T_3)\|=4>\|p\|_\infty$  The idea will be to plug in the matrices $T_2, T_3$ everywhere, and $T_1^3$ for $z_1^3$,  (but leaving $z_1$ alone in the $z_1z_2z_3$ term) to obtain
$$
X= T_1^2+T_2^2+T_3^2 
$$
and 
$$
Y= -T_2T_3.
$$
If we then let $Z=T_1$, we get that $Z$ commutes with $X$ and $Y$ (since all $X, Y, Z$ are all functions of the commuting $T_j$) and
$$
\|X+ZY\|=\|p(T_1, T_2, T_3)\|=4.
$$
We are done if we can show that $\sup_{|z|=1}\|X+zY\|< 4$.  At this point we need to calculate. The Crabb-Davie contractions $T_j$ are constructed as follows: take an orthonormal basis for $\mathbb C^8$ labeled as $e, f_1, f_2, f_3, g_1, g_2, g_3, h$.  For a certain choice of $3\times 3$ unitary matrices $U_1, U_2, U_3$, define
$$
T_j e= f_j \\
T_j f_i =\sum (U_j)_{ik}g_k \\
T_j g_k =h\\
T_j h=0
$$
(I don't have the paper in front of me so this may be slightly off but the idea is the same.) Defined this way, a threefold product $T_iT_jT_k$ has at most one nonzero entry, in the $(h, e)$ position (the "southwest" corner), and the value of this entry is $\langle U_i f_j, g_k\rangle$.  The point of their construction is that the unitaries $U_j$ can be chosen so that
$$
\langle U_i f_j, g_k\rangle =\begin{cases} 1 & \text{ if } i=j=k \\
                                           -1 & \text{ if } i,j,k \text{ all distinct}\\
                                           0 & \text{ otherwise}.\end{cases}
$$
Now, with $X,Y$ defined as above, a little computation shows that the matrix 
$$
X+zY =T_1^2 + T_2^2 + T_3^2 -zT_2T_3
$$
has entries supported only in the first column and last row. 
The last row looks like
$$
(3, zv_1, zv_2, zv_3, 0,0 ,0,0)
$$
where $(v_1,v_2, v_3)$ is a unit vector in $\mathbb C^3$; and the first column looks like
$$
(0,0,0,0,zw_1, zw_2, zw_3, 3)^T
$$
with $3$ being the common entry in the "soutwest" corner, and $w$ another unit vector.
A little more computation shows that for any $|z|\leq 1$, we have $\|X+zY\|\leq \sqrt{13}$, but as $\|X+ZY\|=4$ we have a counterexample. 
