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.  
 A: 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:
$$
\log|z| \le \log^+|x| + \log^+|y| + \log 2.
$$
