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?

Hint : Try factorizing the LHS. Let $f(x) = x^3 + x - 2$, $f(1) = 0$ thus $x - 1$ is a factor of $f$.
$$ f(x) = x^3 + x - 2 = (x - 1)\cdot P(x)$$
where $P(x)$ is a quadratic polynomial of $x$. 
