A set of real numbers $\{\lambda_1, \dots, \lambda_n \}$, $n \geq 1$, is called a Suleĭmanova spectrum if it contains exactly one positive value and $\sum_{i=1}^n \lambda_i \geq 0$. (It is well-known that any such set is the spectrum of an entry-wise $n$-by-$n$ nonnegative matrix).
Let $\{\lambda_1, \dots, \lambda_n \}$ and $\{\hat{\lambda}_1, \dots, \hat{\lambda}_{n+1} \}$ be Suleĭmanova spectra such that
$$\hat{\lambda}_1 \leq \lambda_1 \leq \hat{\lambda}_2 \leq \lambda_2 \leq \dots \leq \hat{\lambda}_n \leq 0 \leq \lambda_n \leq \hat{\lambda}_{n+1}$$
and
$$\sum_{i=1}^{n+1} \hat{\lambda}_i - \sum_{i=1}^n \lambda_i \geq 0.$$
Now let $$ f(t) = \prod_{i=1}^{n+1} (t - \hat{\lambda}_i) $$ and $$ g(t) = \prod_{i=1}^{n} (t - {\lambda}_i).$$
In the first-edition of Horn-and-Johnson's Matrix Analysis, the interlacing-assumption was used to show that $f(\lambda_i)/g'(\lambda_i) \leq 0$, $1\leq i\leq n$ (p. 188).
Let $$ y_i := \begin{cases} -\sqrt{-f(\lambda_i)/g'(\lambda_i)}, & 1\leq i \leq n-1; \\ \sqrt{-f(\lambda_i)/g'(\lambda_i)}, & i=n; \end{cases}$$
Experimental evidence suggests that $\{ y_i \mid 1\leq i \leq n\}$ is a Suleĭmanova spectrum, but I am having difficulty proving that $\sum_{i=1}^n y_i \geq 0$.
