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?

$x^3+x−2=(x−1)(x^2+x+2) \leq0$
Since $(x^2+x+2)> \frac{7}{4}$
$(x-1)\leq0 \Rightarrow x\leq1$.
