Showing $\left|\frac{e^{iz}}{z^2+1}\right|\leq\frac{1}{|z|^2+1}$ 
I wish to show that if $z$ is real, then 
  $$\left|\frac{e^{iz}}{z^2+1}\right|\leq\frac{1}{|z|^2+1}$$

I have shown this result, although my inequality is the wrong way around.
I considered
\begin{align}
|z^2+1|&\leq |z^2|+|1| \ \ \ \ \ \ \ \text{(triangle inequality)} \\
&=|z|^2+1 \\ \\
\Rightarrow |z^2+1|&\leq |z|^2+1 \\
\frac{1}{|z^2+1|}&\geq\frac{1}{|z|^2+1} \\
\frac{|e^{iz}|}{|z^2+1|}&\geq\frac{|e^{iz}|}{|z|^2+1} \\
\left|\frac{e^{iz}}{z^2+1}\right|&\geq\frac{e^{-\text{Im}(z)}}{|z|^2+1} \\
\left|\frac{e^{iz}}{z^2+1}\right|&\geq\frac{1}{|z|^2+1} \ \ \ \ \ \ \text{(if $z$ is real $\Rightarrow$ Im$(z)=0$)} \\
\end{align}
Where did I go wrong?

Also, I wonder, would this inequality still hold if $z$ was not real?


We have
$$\left|\frac{e^{iz}}{z^2+1}\right|=\frac{\left|e^{iz}\right|}{\left|z^2+1\right|}= \frac{1}{\left|z^2+1\right|}$$
and 
$$0\le\left|{z^2+1}\right|= |z|^2+1$$
therefore the result follows.
