How to prove the following inequality of logarithm? Let $x,y,z\in\mathbb{C}.$ Suppose $$z=\frac{1}{2}(xy\pm\sqrt{x^2y^2-4(x^2+y^2)} ).$$
Show that $$log^+|z|\leq log^+|x|+log^+|y|+log 2.$$
Where $log^+\phi=max\{0,log\phi\}.$ Here we are also considering the complex root function with respect to the principle branch of logarithm.  

To prove what you want, we basically just need to apply the triangle inequality to your formula. I'll look at cases separately depending whether $|x|,|y|$ are greater than or smaller than unit magnitude. 


*

*If $|x| \le 1$, $|y| \le 1$, then $|x^2y^2-4x^2-4y^2| \le |x^2y^2|+|4x^2|+|4y^2|\le 9$,so: 
$$2z=xy\pm\left(x^2y^2-4x^2-4y^2\right)^{1/2},$$
$$\begin{aligned}\left|2z\right|&\le|xy|+|x^2y^2-4x^2-4y^2|^{1/2}\\
&\le 1+ 9^{1/2}=4 \end{aligned} $$
Hence $\log\left|z\right| \le \log 2$.

*If $|x| \ge 1$, $|y| \le 1$, then
$$\frac{z}{x}=\frac{y}{2}\pm\left(\frac{y^2}{4}-\frac{y^2}{x^2}-1\right)^{1/2},$$
$$\begin{aligned}\left|\frac{z}{x}\right|&\le\frac{|y|}{2}+\left|\frac{y^2}{4}-\frac{y^2}{x^2}-1\right|^{1/2}\\
&\le \frac{|y|}{2}+ \left(\frac{|y|^2}{4}+\left|\frac{y^2}{x^2}\right|+1\right)^{1/2}  \\
&\le \frac{1}{2}+ \left|\frac{1}{4}+2\right|^{1/2}= \frac{1}{2}+ \left|\frac{9}{4}\right|^{1/2}=2 \end{aligned} $$
Hence $\log\left|\frac{z}{x}\right| \le \log 2$.

*If $|y| \ge 1$, $|x| \le 1$, then by symmetry with $x$ and $y$ and case 2 we have $\log\left|\frac{z}{y}\right| \le \log 2$.

*If $|x| \ge 1$, $|y| \ge 1$, then $|1/x^2+1/y^2| \le 2$ so:
$$\frac{z}{xy}=\frac{1}{2}\pm\left(\frac{1}{4}-\frac{1}{x^2}-\frac{1}{y^2}\right)^{1/2},$$
$$\begin{aligned}\left|\frac{z}{xy}\right|&\le\frac{1}{2}+\left|\frac{1}{4}-\frac{1}{x^2}-\frac{1}{y^2}\right|^{1/2}\\
&\le \frac{1}{2}+ \left(\frac{1}{4}+\left|\frac{1}{x^2}+\frac{1}{y^2}\right|\right)^{1/2}  \\
&\le \frac{1}{2}+ \left|\frac{1}{4}+2\right|^{1/2}= \frac{1}{2}+ \left|\frac{9}{4}\right|^{1/2}=2 \end{aligned} $$
Hence $\log\left|\frac{z}{xy}\right| \le \log 2$.
In summary,
$$
\log|z| \le \begin{cases} \log 2 & \text{ if } |x| \le 1, |y| \le 1\\
\log|x| + \log 2 & \text{ if } |x| \ge 1, |y| \le 1\\
\log|y| + \log 2 & \text{ if } |x| \le 1, |y| \ge 1\\
\log|x| + \log|y| + \log 2 & \text{ if } |x| \ge 1, |y| \ge 1\\
\end{cases}.
$$
Then since $\log^+|x|=\begin{cases}0 & \text{ if } |x| \le 1\\ \log|x| & \text{ if }|x| \ge 1\end{cases}$, this is equivalent to:
$$
\log|z| \le \log^+|x| + \log^+|y| + \log 2.
$$
