Log of Quadratic Form - Mathematical Programming Reformulations Suppose I have the quadratic form 

$$(x-\mu)^{T}Q(x-\mu)$$

where $Q$ is a positive semi-definite matrix. This form achieves a minimum at $x=\mu$. 

Suppose I want to minimize instead

$$\log \left(1 +(x-\mu)^{T}Q(x-\mu) \right)$$

By monotonicity of the $\log$ function this expression, non-convex, also achcieves a minimum at $x=\mu$.

Suppose I didn't know the above and now wanted to solve this via mathematical programming

$$\text{Minimize}_{x\in\mathbb{R}}\quad \log \left(1+(x-\mu)^{T}Q(x-\mu) \right)$$

I could reformulate the problem  in the following way

$$\text{Minimize}_{x\in\mathbb{R},z\in\mathbb{R}}\quad \log \left(z \right)$$
$$\text{s.t.} \quad z\geq 1+(x-\mu)^{T}Q(x-\mu)$$
$$\text{   } \quad z\leq 1+(x-\mu)^{T}Q(x-\mu)$$

The last constraint is not a convex quadratic constraint because the quadratic is bounded from below by $z$.

My question is, isn't this last constraint redundant? Can't I throw it away? My intuition tells me that in order to minimize $\log(z)$ we need to make $z$ as small as possible. $z$ is bounded below by the quadratic form, so we need to make the quadratic form as small as possible. This alone, without the last constraint, would imply that $z=(x-\mu)^{T}Q(x-\mu)$

If I can do away with that last constraint I can do a piecewise linear approximation to the logarithm and solve the problem as a MIQCP.

If I am correct, how can I formally argue the reformulation using only the first quadratic constraint?
I'm not sure that "equivalence" between optimization models is all that well defined. If we think of a problem instance as the combination of a model (with symbolic parameters) and values for the parameters (data), then one might say two models are equivalent for a problem if, given the same data, any optimal solution to either model is also optimal in the other.

Using that definition, let (P) denote the final model (with both inequalities) and (P') the model with the second inequality removed. To show (P') is equivalent to (P), you don't need to show the feasible regions are the same, just that any optimal solution to (P) is optimal in (P') and vice versa. It's easy to prove that $(x, z)$ optimal in (P') implies $z = (x-\mu)^TQ(x-\mu)$, after which the equivalence proof is pretty trivial.
The equivalent is

$$
\begin{cases}
z^*=\text{argmin} \{\log(z)\} \\
\text{st.}\quad z\ge1
\end{cases}
$$

Which does not look like a multi-variable programming anymore.

The solution is $z=1$, which means

$$z=1 +(x-\mu)^{T}Q(x-\mu)=1$$

or

$$x=\mu$$