Showing moderate decrease property in the real line for $f(z)=\frac{a}{a^2+z^2}$. Let $f(z) = \frac{a}{a^2 + z^2}$ for $a>0$. Then $f$ is holomorphic in the horizontal strip 
$|\Im(z)| < a$. I would like to show that in if $|\Im (z)| < a/2$, we have some constant $A>0$ such that
$$|f(x+iy)| \le \frac{A}{1+x^2}$$ for all $x \in \mathbb{R}$ and $|y|<a/2$. 
To get this bound, first note that $f(x+iy) = \frac{a}{a^2 + x^2 -y^2 + 2xy i}$. 
So when we take the absolute value, we need a bound like $|a^2+x^2 - y^2+2xy i | \ge 1+x^2$. 
I can see that we have $|a^2+x^2-y^2 + 2xyi| \ge |a^2+x^2-y^2| \ge (a^2+x^2) - y^2 > \frac{3a^2}{4} + x^2$. 
So we have $|f(x+iy)| \le \frac{a}{\frac{3a^2}{4} + x^2}$. But how can I bound this right fraction by some $\frac{A}{1+x^2}$? 

All you need is the following: If $c>0,$ then
$$\tag 1 \frac{1+x^2}{c+x^2}\, \text{is bounded on }\mathbb R.$$
Can you prove this?
