How to show the following inequality with f(x) = 1/(1+x^2)?

Question

How to show the following inequality : Let, $f(x) = \frac{1}{1+x^2}$ . Then show that $$\int_{\mathbb{R}\setminus (-1,1)} \left( \sqrt{f(x+y)}-\sqrt{f(x)}\ \right)^2 \ \frac{dy}{y^2} \leq C f(x) $$
We can choose $x\geq 0$ . Actually if we devide the integral by $\int_{-\infty} ^{-1} + \int^{\infty} _{1} $ we can easily see that : $$\int^{\infty} _{1} \left( \sqrt{f(x+y)}-\sqrt{f(x)}\ \right)^2 \ \frac{dy}{y^2} \\ \leq C\int^{\infty} _{1}f(x+y) \frac{dy}{y^2} +C f(x) \int^{\infty} _{1} \frac{dy}{y^2} \\ \leq C f(x). $$ But I am not able to prove the other part. Or there might be a better way to do that .