I am reading this paper which resolves the famous Kadison-Singer problem. Recalling Definition 5.3 We say that $z\in R^m$ is above the roots of the polynomial $p(x_1,\cdots, x_m)$ (denoted as $z\in Ab_p$) if $p(z+t)>0,\forall t\in R^m_+$.
Now consider a univariate real rooted polynomial $p$, with positive leading coefficient. If $y>maxroot(p)$, then clearly $y\in Ab_p$, because $p(y+t) = \prod_i(y-\lambda_i+t) > 0, \forall t \geq 0$.
It turns out that the converse is also true(see the paragraph following the definition, where this fact is used, see also this, section 8.3, page 36). How to prove this? My reasoning is the following:
Let $\lambda_1\geq\cdots\geq\lambda_n$ are the roots of $p$. After appropriate scaling we can assume that the leading coefficient is 1, so
$p(x) = \prod_{i=1}^n(x-\lambda_i)$. Let $y\in Ab_p$. This means
$p(y+t) = \prod_{i=1}^n(y-\lambda_i + t) > 0, \forall t\geq 0$
Now suppose that $\exists r\in[n]$ such that $\lambda_1\geq\cdots\lambda_r>y>\lambda_{r+1}\geq\lambda_n$. Now the first $r$ term in the factorisation, corresponding to the largest $r$ root is of the form $(-ve +t)$, and the remaining $n-r$ terms are of the form $(+ve + t)$. Now it is clear that is we choose sufficiently small non negative number as $t$, then $p(y+t)<0$, when $r$ is odd, a contradiction.
But the above argument does not hold true when $r$ is even.
If $r$ is even, then $r\ge 2$, so $\lambda_2>y$. Now you can find $t$ such that $\lambda_1 >y+t>\lambda_2$ so that $p(y+t)<0$.