Show that $\lvert \cos z \rvert \leqslant \cosh \lvert z \rvert$ for all $z \in \mathbb{C}$. I am trying to prove the inequality
$$\lvert \cos z \rvert \leqslant \cosh \lvert z \rvert.$$
So far, I have (writing $z = a + bi$ with $a, b \in \mathbb{R})$
$$ \lvert \cos z \rvert = \lvert{\frac{e^{ia} e^{-b} + e^{-ia} e^{b}}{2}}\rvert \leqslant \frac{\lvert{e^{ia} e^{-b}\rvert} + \lvert{e^{-ia} e^{b}}\rvert}{2} = \frac{e^{-b} + e^{b}}{2}, $$
by the triangle inequality and the fact that $e^{ia}$ is on the unit circle. I also have
$$ \cosh \lvert z \rvert = \cos(i\lvert z \rvert) = \frac{e^{-\lvert z \rvert} + e^{\lvert z \rvert}}{2} = \frac{e^{-\sqrt{a^2+b^2}} + e^{\sqrt{a^2+b^2}}}{2}.$$
Thus, it suffices to prove
$$ \frac{e^{-b} + e^{b}}{2} \leqslant \frac{e^{-\sqrt{a^2+b^2}} + e^{\sqrt{a^2+b^2}}}{2}.$$
I don't seem to be able to prove this, nor does it seem to be true in general. Is this even the right way to go about this problem?
(Note: in the original post, I mistakenly wrote $\lvert z \rvert = a^2 + b^2$)
 A: You solved most of the problem. But you need to correct  $|z|=\sqrt{a^2+b^2}$ everywhere through.
The last step then is:
Since the function $x\in \mathbb{R}\mapsto \cosh{x}$ is strictly increasing on $[0,\infty[$
and since $|b|\leq \sqrt{a^2+b^2}, \; \forall a,b \in \mathbb{R}$, then
$$\cosh{|b|}< \cosh{\sqrt{a^2+b^2}}.$$
And since $\forall b\in \mathbb{R}, \cosh{b}=\cosh{|b|}$ becasue the hyperbolic cosine is an even function, then
$$\cosh{b}< \cosh{\sqrt{a^2+b^2}}$$
which is your last (and required) inequality.
A: Note that if $z=a+ib$, one has $\cos z=\cos a \cosh b - i \sin a \sinh b$ so $|\cos z|^2=\cos^2a +\sinh^2 b$
(using $\cosh^2 b=1+\sinh^2 b$)
On the other hand:
$\cosh^2 |z|=\cosh^2 \sqrt {a^2+b^2} \ge \cosh^2 |b| = 1 + \sinh^2 |b| \ge  \cos^2 a +\sinh^2 |b|=|\cos z|^2$
A: Big, big hint: replace each of $\cos z$ and $\cosh |z|$ by their respective power series expansions:
\begin{align*}
\cos z = \sum_{k = 0}^\infty \frac{(-z^2)^{k}}{(2k)!} &= 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + ... \\
\cosh |z| = \sum_{k = 0}^\infty \frac{|z^2|^{k}}{(2k)!} &= 1 + \frac{|z|^2}{2!} + \frac{|z|^4}{4!} + \frac{|z|^6}{6!} + ... \
\end{align*}
and sub into the given inequality.
You can do it without power series, but to my thinking the solution is by far clearest and most intuitive--in fact, almost trivially obvious--if you do it this way.
