# Determine all real numbers $x$ that satisfy the inequality

Let $$r$$ be a fixed real number. Determine all real numbers $$x$$ that satisfy the inequality $$\frac{1}{1+x^2}$$ $$≤ r$$ .

Can someone help me start this question? I am aware of a method when $$r = 0$$ , but I don't know how to do this since $$r$$ is unknown.

Note that, since $$1+x^2>0$$, for $$r\le 0$$ there are not solutions then we can consider the case $$r>0$$ and obtain $$\frac{1}{1+x^2}\le r \iff (1+x^2)\frac{1}{1+x^2}\le (1+x^2)r \iff rx^2+r-1\ge 0$$
• @Taylor Just isolate the $x^2$ term. Show your work editing the OP, I'll take a look to it giving some more suggestion if you'll need it. – gimusi Nov 15 '18 at 17:48
• Another way to isolate $x^2$ is by inverting both sides of the inequality: $1+x^2 \ge \frac{1}{r}$. This is allowed as long as $r \ne 0$. – Cuspy Code Nov 15 '18 at 17:53