Let $t_1,t_2,\ldots,t_n$ be all roots of $P(x)$, where $t_i$s are in $\mathbb{R}^+$, and not necessarily distinct numbers. So we have that $a_{n-1}=\sum_{i=1}^nt_i=\sum t_i$ and $a_{n-2}=\sum_{1\le\i<j\le n}t_it_j=\sum t_it_j$. From Symmetric Means Inequality we have $\bigg{(} \frac{\sum t_it_j}{\binom{n}{2}}\bigg{)}^{\frac{1}{2}}\le \frac{\sum t_i}{n} \implies \frac{2a_{n-2}}{n(n-1)}\le \frac{a_{n-1}^2}{n^2}\implies \frac{a_{n-1}^2}{a_{n-2}}\geq \frac{2n}{n-1}$. So minimum value of $ \frac{a_{n-1}^2}{a_{n-2}}$ is $\frac{2n}{n-1}$ and equality occurs when all $t_i$ are equal to each others.

Jalil_Huseynov wrote: Let $t_1,t_2,\ldots,t_n$ be all roots of $P(x)$, where $t_i$s are in $\mathbb{R}^+$, and not necessarily distinct numbers. So we have that $a_{n-1}=\sum_{i=1}^nt_i=\sum t_i$ and $a_{n-2}=\sum_{1\le\i<j\le n}t_it_j=\sum t_it_j$. From Symmetric Means Inequality we have $\bigg{(} \frac{\sum t_it_j}{\binom{n}{2}}\bigg{)}^{\frac{1}{2}}\le \frac{\sum t_i}{n} \implies \frac{2a_{n-2}}{n(n-1)}\le \frac{a_{n-1}^2}{n^2}\implies \frac{a_{n-1}^2}{a_{n-2}}\geq \frac{2n}{n-1}$. So minimum value of $ \frac{a_{n-1}^2}{a_{n-2}}$ is $\frac{2n}{n-1}$ and equality occurs when all $t_i$ are equal to each others. bravo