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?


If $z$ is real: $|e^{iz}|=1$, $|z^2+1|=z^2+1=|z|^2+1$, so that
$$\left|\frac{e^{iz}}{z^2+1}\right|=\frac{1}{|z|^2+1}.$$
If $z=x+iy$ is complex, with $y$ a large negative number, then $|e^{iz}|
=e^{-y}$ is huge, and so
$$\left|\frac{e^{iz}}{z^2+1}\right|\gg\frac{1}{|z|^2+1}.$$
