Normal-defect inequality for cyclic matrices of consecutive sizes

Let $A \in \mathbb{C}^{n\times n}$ and $\tilde{A}\in \mathbb{C}^{(n+1)\times (n+1)}$ with $n \geq 4$ be the matrices defined by

$$A = \begin{pmatrix} 0 & a_1 & & \cdots & 0 \\ & & a_2 & & \\ \vdots & & & \ddots & \vdots \\ & & & & a_{n-1} \\ a_n & & & \cdots & 0 \end{pmatrix},\qquad \tilde{A} = \begin{pmatrix} 0 & a_1 & & \cdots & & 0 \\ & & a_2 & & \\ & & & \ddots & & \vdots \\ & & & & a_{n-1} & \\ & & & & & a_n \\ 0 & & & \cdots & & 0 \end{pmatrix}.$$

Here $\operatorname{nd}$ denotes the normal defect. Normal-defect inequality. Then

$$\operatorname{nd}(\tilde{A}) \leq \operatorname{nd}(A)+1.$$

The inequality is motivated by the fact that a minimal normal completion of $A$ can be used to construct a normal completion of $\tilde{A}$ with one additional completion dimension; whether the resulting bound is always valid remains open.