Estonian Mathematical Olympiad

a. Let $c$ be any root of $P(x)$. As $P(0) = a_0 = a_n \neq 0$, we must have $c \neq 0$. Notice that
$$\begin{aligned} P\left(\frac{1}{c}\right) &= a_n \left(\frac{1}{c}\right)^n + a_{n-1} \left(\frac{1}{c}\right)^{n-1} + \dots + a_1 \left(\frac{1}{c}\right) + a_0 \\ &= \left(\frac{1}{c}\right)^n (a_0 c^n + a_1 c^{n-1} + \dots + a_{n-1} c + a_n) \\ &= \left(\frac{1}{c}\right)^n (a_n c^n + a_{n-1} c^{n-1} + \dots + a_1 c + a_0) \\ &= \left(\frac{1}{c}\right)^n P(c) = 0, \end{aligned}$$
which shows that $\frac{1}{c}$ is also a root of $P(x)$. Thus the roots can be divided into inverse pairs.
For each root $c$ we have $|c| + \left|\frac{1}{c}\right| \ge 2$. The only roots paired with itself are 1 and -1, both of which have an absolute value of 1. If there are $l$ such roots, then
$$|x_1| + |x_2| + \dots + |x_k| \ge \frac{k-l}{2} \cdot 2 + l \cdot 1 = k - l + l = k.$$

b. The polynomial $P(x) = x^3 - x^2 - x + 1 = (x-1)^2(x+1)$ satisfies the conditions of the problem and its real roots are $x_1 = 1$ and $x_2 = -1$. Thus $k = 2 > 1$, but $|x_1| + |x_2| = 2 = k$, so the inequality is non-strict.