Solving degree three inequality 
$$\frac{2}{(x^2+1)}\geq x.$$

I thought to do this:$$2-x^3-x\geq 0$$
But we haven't learnt how to solve equations of third grade. Could you help me maybe by factorizing sth?

\begin{align}
\frac{2}{x^2+1} &\ge x \iff \\
2 &\ge x(x^2+1) = x^3 + x \iff \\
0 &\ge x^3 + x - 2
\end{align}
Trying some easy roots one gets $x=1$:
\begin{align}
0 &\ge (x-1)(x^2+x+2) \quad (*)
\end{align}
Then we have
$$
x^2 +x+ 2 = (x+1/2)^2 + 7/4 \ge 7/4
$$
which means no real roots for this part, no change of sign. So we can divide both sides by the above and simplify $(*)$ to:
\begin{align}
0 &\ge x - 1 \iff \\
1 &\ge x
\end{align}
