Let $P(x)=a_nx^n+...+a_1x+a_0$ be a polynomial with $x_0$ being a real root of $P(x)$.
If $P(x)=x-1$, we have $x_0=1$ and $|x_0|=1 \leq 1+|\frac{-1}{1}|=2$
If $P(x)=x^2+2x+1$, we have $x_0=-1$ and $|x_0|=1 \leq 1+|\frac{2}{1}|=2$
If $P(x)=2x^2-5x+2$, if $x_0=2$ then $|x_0|=2 \leq 1+|\frac{-5}{2}|=\frac{7}{2}$, if $x_0=\frac{1}{2}$ then $|x_0|= \frac{1}{2}\leq 1+|\frac{-5}{2}|=\frac{7}{2}$
Is it true that for every real root $x_0$ of $P(x)$, there exist an number $i$ such that $x_0 \leq 1+|\frac{a_i}{a_n}|$ ?
Yes, this is the so called Cauchy bound. (see for instance https://en.wikipedia.org/wiki/Properties_of_polynomial_roots#Other_bounds and https://captainblack.wordpress.com/2009/03/08/cauchys-upper-bound-for-the-roots-of-a-polynomial/ for a proof.)