Assume $1/4 \leqslant\rho < 1$. Why does
$\displaystyle \frac{1}{\pi}\left| \log \left| \frac{1 + \rho e^{i\theta}}{1 - \rho e^{i\theta}}\right|\right| \leqslant \frac{1}{\pi} \left( \log 4 + \log \frac{1}{\left| \cos(\theta/2)\right|} + \log\frac{1}{\left| \sin(\theta/2)\right|}\right)$
hold?
Here is the proof my professor stated. We have on the one hand
$\left| 1 + \zeta \right| \leqslant 1 + \left| \zeta \right| = 1 + \rho$
and on the other hand
$\left| 1 - \zeta \right| \geqslant \left|\mathrm{Im}\left(1 - \zeta\right) \right| = \rho\left|\sin(\theta)\right| $
Hence
$\displaystyle \log \frac{\left| 1 + \zeta \right|}{\left| 1 - \zeta \right|}\\ \leqslant \log\left(\frac{1 + \rho}{\rho \left| \sin(\theta)\right|}\right) \\ = \log\left(\frac{1 + \rho}{2\rho \left| \sin(\theta/2)\right|\left| \cos(\theta/2)\right|} \right)\\ = \log\left(\frac{1 + \rho}{2\rho}\right) + \log \left(\frac{1}{\left| \sin(\theta/2)\right|}\right) + \log\left( \frac{1}{\left| \cos(\theta/2)\right|} \right)\\ \leqslant \log(5/2) + \log \left(\frac{1}{\left| \sin(\theta/2)\right|}\right) + \log\left( \frac{1}{\left| \cos(\theta/2)\right|} \right)$
Since $(1 + \rho)/(2\rho) = 1/2 + 1/(2\rho) \leqslant 5/2$ by $\rho \geqslant 1/4$. But I am not sure how to proceed if the absolute value of the logarithm is invoked.
The way OP split the expression into those two parts and apply the algebraic manipulation as such is on the right track, but some tiny things involving absolute values are missed here and there. For this very reason, I shall be as detailed as I can.
Not that it affects the final inequality, but $\ldots$ theres's an error in the half part OP already figured out: making this inequality $$\text{L.H.S.}\equiv 1+\rho^2 +2\rho\cos\theta < 2 + 2\cos\theta \equiv\text{R.H.S.}$$ just intuitively from $\rho < 1$ is simply wrong.
Plot the curve or solve $\text{L.H.S.} = \text{R.H.S.}$ you'll see that the inequality is reversed at the critical point $\rho = -1 -2\cos\theta$. For example, given $\rho = -1 +\sqrt{2}$ you have L.H.S. > R.H.S. for $\frac{3\pi}{4} < \theta < \frac{5\pi}{4}$.
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There's nothing particularly advanced here, just that the process is tedious.
The two inequalities invoked a lot are $$|x + y| \leq |x| + |y| \qquad \text{and} \qquad |x - y| \leq |x| + |y|$$ both with plus on the right. I will quote either of them as the `tri-inEq' (triangle inequality).
When you plot this thing you'll see that for a given $\rho$ the expression as a function of $\theta$ is monotonic decreasing from zero to $\pi/2$, and after the discontinuity in derivative at $\theta = \pi/2$, it is a `mirror image' of the left. The whole thing repeats with a period of $\pi$.
As shown in the figure (where $\rho = 0.98$), the blue curve is the original expression and the yellow curve is the desired upper bound. The horizontal green dashed line is at a height of $\log 8 = 3\log 2$.
Hence, it suffices to consider only the open interval $\theta \in (0,\, \frac{\pi}2)$
$$ \begin{align}A \equiv \left|~ \log\left| \frac{ 1 + \rho e^{i\theta} }{ 1 - \rho e^{i\theta} } \right| ~\right| &= \left|~ \log \frac{ \left|1 + \rho e^{i\theta} \right|}{ \left|1 - \rho e^{i\theta} \right| } ~\right| \\ &=\left|~~ \log \left|1 + \rho e^{i\theta} \right| - \log \left|1 - \rho e^{i\theta} \right| ~~~\right| \\ &\leq \left|~ \log \left|1 + \rho e^{i\theta} \right| ~\right| + \left|~\log \left|1 - \rho e^{i\theta} \right| ~\right| \tag{tri-inEq} \end{align} $$
Denote the split as $A = A_1 + A_2$ , where
$$ \begin{align} A_1 &\equiv \left|~ \log \left|1 + \rho e^{i\theta} \right| ~\right| = \left|~ \log \sqrt{1 + \rho^2 + 2\rho \cos\theta} ~\right|\\ &= \left|~~~ \log\sqrt{1 + \rho^2} ~~ + \log\sqrt{1 + \frac{2\rho}{1 + \rho^2}\cos\theta } ~~~\right| \\ &\leq \left|~ \log\sqrt{1 + \rho^2} ~\right| + \left|~ \log\sqrt{1 + \frac{2\rho}{1 + \rho^2}\cos\theta } ~~~\right| \tag{tri-inEq} \\ &\leq \left|~ \log\sqrt{1 + \rho^2} ~\right| + \left|~ \log\sqrt{1 + \cos\theta } ~~\right| \end{align}$$ because $2\rho \leq 1 + \rho^2 ~~ \forall \rho \in \mathbb{R}$, the square root function is monotone increasing as is the logarithm function, and cosine is positive in this region $\theta \in (0, \pi/2)$ thus the argument $(1 + \frac{2 \rho }{1 + \rho^2}\cos\theta)$ in log is over unity (making log positive) and growing. $$ \begin{align} &< \left|~ \log\sqrt{2} ~\right| + \left|~ \log\sqrt{1 + \cos\theta } ~~\right| \tag*{$\because \rho < 1$} \\ &= \log\sqrt{2} ~~ + \left|~ \log\sqrt{2} ~ + \log\sqrt{\frac{ 1 + \cos\theta}2 } ~~\right| \\ &\leq \log\sqrt{2} + \left| \log\sqrt{2} \right| ~ + \left|~ \vphantom{\frac{x}2}\log\left| \cos(\theta/2) \right| ~\right| \tag{tri-inEq}\\ &= \log 2 + \log\frac{1}{\left| \cos(\theta/2) \right|} \tag*{$\because \cos(\theta/2) \leq 1$ always} \end{align}$$
Similarly for the other half OP asked about:
$$ \begin{align} A_2 &\equiv \left|~ \log \left|1 - \rho e^{i\theta} \right| ~\right| = \left|~ \log \sqrt{1 + \rho^2 - 2\rho \cos\theta} ~\right|\\ &= \left|~~~ \log\sqrt{1 + \rho^2} ~~ + \log\sqrt{1 - \frac{2\rho}{1 + \rho^2}\cos\theta } ~~~\right| \\ &\leq \left|~ \log\sqrt{1 + \rho^2} ~\right| + \left|~ \log\sqrt{1 - \frac{2\rho}{1 + \rho^2}\cos\theta } ~~~\right| \tag{tri-inEq} \\ &\leq \left|~ \log\sqrt{1 + \rho^2} ~\right| + \left|~ \log\sqrt{1 - \cos\theta } ~~\right| \end{align}$$ this time because $2\rho \leq 1 + \rho^2$ and cosine is negative, the argument in log is SMALLER than unity (making log negative), and $(1 - \frac{2 \rho }{1 + \rho^2}\cos\theta) \mapsto (1 - \cos\theta)$ is making logarithm more negative thus growing in absolute value. $$ \begin{align} &< \left|~ \log\sqrt{2} ~\right| + \left|~ \log\sqrt{1 - \cos\theta } ~~\right| \tag*{$\because \rho < 1$} \\ &= \log\sqrt{2} ~~ + \left|~ \log\sqrt{2} ~ + \log\sqrt{\frac{ 1 - \cos\theta}2 } ~~\right| \\ &\leq \log\sqrt{2} + \left| \log\sqrt{2} \right| ~ + \left|~ \vphantom{\frac{x}2} \log\left| \sin(\theta/2) \right| ~\right| \tag{tri-inEq}\\ &= \log 2 + \log\frac{1}{\left| \sin(\theta/2) \right|} \tag*{$\because \sin(\theta/2) \leq 1$ always} \end{align}$$
Thus finally, $$A = A_1 + A_2 = 2\log 2 + \log\frac{1}{\left| \cos(\theta/2) \right|} + \log\frac{1}{\left| \sin(\theta/2) \right|}$$ is the desired estimate.
The cumbersome steps in bounding (pushing the inequalities) might be necessary in some sense. The absolute value works on two fronts: at zero for general functions, and at one for logarithm ... if you make the plots of the stages of functions from $| 1 + \rho^2 \pm \cos\theta|$ to the final expression, you see a complicated `folding' process.
One have already seen in the figure that this upper bound is pretty loose for all angles except around $\theta \approx k\pi,\, k \in \mathbb{Z}$ for $\rho \approx 1$. Around these singularities, the estimate can be thought of as better OR worse than ordinary regions, depending on how one measures the error for the $\rho \approx 1$ case. Just for the record below is a figure for $\rho = 0.600$
p.s.
I wonder what the overall factor $\frac{1}{\pi}$ is about? Does this expression come from a contour integral?