Well, if the roots are $x_1,x_2,\dots,x_n$ then we have $x_1+x_2+\dots+x_n=0$ and $a_{n-2}=\sum x_ix_j$ Obviously $(n-2)\sum_{2}^n x_i^2-2\sum_{2\le i<j\le n}x_ix_j=\sum_{2\le i<j\le n} (x_i-x_j)^2\ge0$, or $n\sum_2^nx_i^2+2\sum_{2\le i<j\le n} x_ix_j\ge2(\sum_2^n x_i)^2=-2x_1(\sum_2^nx_i)$, which is equivalent to $2(1-n)\sum_{1\le i<j\le n} x_ix_j\ge nx_1^2$, which is what we wanted. Q.E.D.

Here is a different (and also easy) approach than the one above.

EDIT: Dang OK. Wait actually could you explain how this is bad?

But that only proves that there is some root for which the inequality holds. You're supposed to prove it for all roots.

Let the roots of $f$ be $r_1, r_2, ..., r_n$ where $r_i$ is real. The information that we extract from the polynomial is that $\displaystyle\sum_{i = 1}^n r_i = 0$. From Vieta: $$a_{n-2} = \sum_{1 \le i < j \le n} r_ir_j$$ So we want to show that \begin{align*} |r_k| &\le \sqrt{\frac{2(1-n)}n \sum_{1 \le i < j \le n} r_ir_j} \\ r_k^2 &\le \frac{2(1-n)}n \sum_{1 \le i < j \le n} r_ir_j \end{align*} Notice that $(r_1+\cdots + r_n)^2 = (r_1^2 + \cdots + r_n^2) + 2a_{n-2}$, so our inequality reduces down to $$r_k^2 \le \frac{n-1}n \sum_{i=1}^n r_i^2$$ This becomes \begin{align*} r_k^2 &\le (n-1) \sum_{i = 1, i \neq k}^n r_i^2 \\ \left( \sum_{i = 1, i \neq k}^n r_i \right)^2 &\le (n-1) \sum_{i = 1, i \neq k}^n r_i^2 \end{align*} which directly follows from Cauchy-Schwarz.

PlatinumFalcon wrote: Assume for the sake of contradiction that $x_i^2>\frac{2(1-n)}{n}a_{n-2}=\left(\frac{n-1}{n}\right)S$ for all $i$. Summing over all $i$, we get $S=\sum x_i^2>n\cdot\left(\frac{n-1}{n}\right)S=(n-1)S$, or $(2-n)S>0$, contradiction, as $n\ge 2$. The error is right there; the contradictory assumption doesn't quite work out. It should be "Assume for the sake of contradiction that there exists and $x_i$ such that $x_i^2 > \frac{2(1-n)}{n}a_{n-2}$. The problem says to prove that every root of the polynomial satisfies the inequality, so the negation of the statement would be to assume that there exists a root that doesn't satisfy the bound.

Uhm. How annoying. Let $r_1,r_2,\dots,r_n$ be the roots of $f$. By Vieta $r_1+r_2+\dots+r_n=0$ and $\displaystyle{a_{n-2}=\sum_{1 \leq i < j \leq n} r_ir_j}$. For $n=2$ the inequality is easy. Now for $n \geq 3$ let $1 \leq m \leq n$. We'll show the inequality for $|r_m|$. Basically plug in $|r_m|=\left| \sum_{\substack{i \neq m \\ 1 \leq i \leq n}} r_i \right|$ and represent $a_{n-2}$ with the roots as above via Vieta. In that form substitute $r_m$ again and after some semi-tedious calculations, we'll get \[ 2 \sum_{\substack{i,j \neq m \\ 1 \leq i < j \leq n}} r_ir_j \leq (n-2) \sum_{\substack{i \neq m \\ 1 \leq i \leq n}} r_i^2 = \sum_{\substack{i,j \neq m \\ 1 \leq i < j \leq n}} \left(r_i^2+r_j^2 \right), \]which is immediate via AM-GM. $\hfill \blacksquare$

Let the roots be $r_1, \ldots, r_n$. By Vieta's we obtain $\sum_j r_j^2 = -2a_{n-2}$. Note by Cauchy for all $i$, \[ \left(\sum_{j \neq i} r_j^2\right)(n-1) \ge \left(\sum_{j \neq i} r_j\right)^2 = r_i^2 \implies (n-1)\sum_{j \neq i} r_j^2 \ge r_i^2 \implies (n-1) \sum_j r_j^2 \ge nr_i^2 \]Thus for all $i$ \[ r_i^2 \le \frac{n-1}{n} \sum_j r_j^2 = \frac{-2(n-1)}{n}a_{n-2} \]as wanted.