Estonian Mathematical Olympiad

a. Let $c$ be any root of $P(x)$. As $P(0)=a_0=a_n\ne 0$, we must have $c\ne 0$. Notice that $$\begin{array}{rl}P\left(\frac{1}{c}\right)& =a_n{\left(\frac{1}{c}\right)}^n+a_{n-1}{\left(\frac{1}{c}\right)}^{n-1}+\cdots +a_1\left(\frac{1}{c}\right)+a_0\\ & ={\left(\frac{1}{c}\right)}^n(a_0{c}^n+a_1{c}^{n-1}+\cdots +a_{n-1}c+a_n)\\ & ={\left(\frac{1}{c}\right)}^n(a_n{c}^n+a_{n-1}{c}^{n-1}+\cdots +a_{1}c+a_0)\\ & ={\left(\frac{1}{c}\right)}^{n}P(c)=0,\end{array}$$ 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 $$|x_1|+|x_2|+\cdots +|x_k|\ge \frac{k-l}{2}\cdot 2+l\cdot 1=k-l+l=k.$$

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.