Let $r_1, r_2, \cdots, r_n$ be the roots of $f(x)=0$. It is clear that none of them will be zero. Therefore, it is possible to divide both sides of this equation by $x^n$ and then let $y=1/x$. $$g(y)=y^n + y^{n-1}+y^{n-2}+a_{n-3}y^{n-3}+\cdots + a_{n-1}y+a_n=0$$ It follows that $1/r_i, (i=1, 2, \cdots, n)$ are all the roots of $g(y)=0$. By the Vieta's theorem, $$\sum_{i=1}^n\Big(\frac{1}{r_i}\Big)=-1\quad\text{and}\quad\sum_{1\le i\le j\le n}\Big(\frac{1}{r_i}\cdot\frac{1}{r_j}\Big)=1$$ $$\therefore\quad\sum_{i=1}^n\Big(\frac{1}{r_i}\Big)^2=\Big(\sum_{i=1}^n\frac{1}{r_i}\Big)^2-2\sum_{1\le i\le j\le n}\Big(\frac{1}{r_i}\cdot\frac{1}{r_j}\Big)=(-1)^2-2\times 1 =-1$$ This means that $r_i$ cannot be all real.