For which real numbers $x$ does the following inequality hold?

$$
x^{6}+4 x^{5}+2 x^{4}-6 x^{3}-2 x^{2}+4 x-1 \geq 0 .
$$
The expression on the left side has a value of -1 at $x=0$ and +2 at $x=1$. Thus, the inequality to be solved is not an identity, and we need to determine the roots of the left side. There, the

$$
f(x)=x^{6}+4 x^{5}+2 x^{4}-6 x^{3}-2 x^{2}+4 x-1
$$

is an integer-coefficient polynomial, and its rational roots can only be numbers of the form $p / q$, where $p$ is a divisor of the constant term, and $q$ is a divisor of the leading coefficient. Since in our case the absolute value of both numbers is 1, the only possible rational roots are -1 and +1. However, these are not roots: $f(-1)=-2$ and $f(1)=2$.

Thus, we need to try to factorize the polynomial. Notice that the absolute values of the coefficients are symmetric around the middle term, and the signs of the pairs alternate or are the same. Polynomials of this kind are often called reciprocal polynomials, and finding their roots is often facilitated by introducing the new variable $y=x-1 / x$. Let's do that! Let $x \neq 0$ and factor out $x^{3}$ from the polynomial:

$$
\begin{gathered}
f(x)=x^{3}\left(\left(x^{3}-\frac{1}{x^{3}}\right)+4\left(x^{2}+\frac{1}{x^{2}}\right)+2\left(x-\frac{1}{x}\right)-6\right)= \\
=x^{3}\left(\left(x^{3}-\frac{1}{x}\right)^{3}+3\left(x-\frac{1}{x}\right)+4\left(x-\frac{1}{x}\right)^{2}+8+2\left(x-\frac{1}{x}\right)-6\right)= \\
=x^{3}\left(y^{3}+4 y^{2}+5 y+2\right)
\end{gathered}
$$

We try to factorize the second factor. We notice that the expression $y^{3}+4 y^{2}+5 y+2$ vanishes at $y=-1$, so $(y+1)$ can be factored out:

$$
y^{3}+4 y^{2}+5 y+2=(y+1)\left(y^{2}+3 y+2\right)=(y+1)(y+1)(y+2)
$$

Substituting this back into the above form of $f(x)$,

$$
f(x)=x^{3}(y+1)^{2}(y+2)=x^{2}\left(x-\frac{1}{x}+1\right)^{2} \cdot x\left(x-\frac{1}{x}+2\right)=\left(x^{2}+x-1\right)^{2}\left(x^{2}+2 x-1\right)
$$

The left and right sides match for $x \neq 0$ due to the equivalence of the transformations, and for $x=0$, $f(0)=-1$, and the value of the right side is also -1. We have obtained the desired factorization, so the inequality to be solved is of the form:

$$
\left(x^{2}+x-1\right)^{2}\left(x^{2}+2 x-1\right) \geq 0
$$

The first factor is the square of a real number, so it is certainly non-negative. Thus, the inequality holds if either

$$
x^{2}+x-1=0
$$

or

$$
x^{2}+2 x-1 \geq 0
$$

The first case is satisfied for $x_{1}=(-1-\sqrt{5}) / 2$ and $x_{2}=(-1+\sqrt{5}) / 2$, and the second case is satisfied if $x \leq-1-\sqrt{2}$ or $x \geq-1+\sqrt{2}$. The latter includes $x_{2}$, since $x_{2}>-1+\sqrt{2}$.

In summary, the inequality is satisfied precisely when

$$
x \leq-1-\sqrt{2}, \quad \text { or } \quad x=\frac{-1-\sqrt{5}}{2}, \quad \text { or } \quad x \geq-1+\sqrt{2}
$$

Remark. Many solvers reasoned after the factorization that the square of a number is non-negative, so the inequality is equivalent to the inequality $x^{2}+2 x-1 \geq 0$. This line of thought is flawed, and such solutions were awarded 3 points.