I want to bound the difference between two logarithmic expression shown below with a constant number i.e not function of $x,y,z$ where $x,y,z \in \mathbb{C}$. The difference is
$$ \log\bigl(1+(|z|+|x|+|y|)^2\bigl)-\log \bigl(1 + (|z| +\sqrt{|x|^2+|y|^2})^2\big) \leq \text{CONSTANT} $$
Any ideas is much appreciated.
Replacing $|x|,|y|,|z|$ by $a,b,c$ and exponentiating, we need to show that $$ \frac{1+(a+b+c)^2}{1+(a+\sqrt{b^2+c^2})^2} \le \text{CONSTANT} $$ for nonnegative real numbers $a,b,c$. This is trivially true in the compact region $\{ (a,b,c)\in\Bbb R_{\ge0}^3\colon a+\sqrt{b^2+c^2} \le 1 \}$. Outside this region, we have $a+b+c \ge a+\sqrt{b^2+c^2} \ge 1$, which implies $1+(a+b+c)^2 \le 2(a+b+c)^2$; it therefore suffices to show that $$ \frac{a+b+c}{a+\sqrt{b^2+c^2}} \le \text{CONSTANT}, $$ since $$ \frac{1+(a+b+c)^2}{1+(a+\sqrt{b^2+c^2})^2} \le \frac{2(a+b+c)^2}{(a+\sqrt{b^2+c^2})^2} = 2 \bigg( \frac{a+b+c}{a+\sqrt{b^2+c^2}} \bigg)^2. $$ But since $b+c \le 2\max\{b,c\}$ while $\sqrt{b^2+c^2} \ge \max\{b,c\}$, we see (setting $m=\max\{b,c\}$) that $$ \frac{a+b+c}{a+\sqrt{b^2+c^2}} \le \frac{a+2m}{a+m} \le \frac{2a+2m}{a+m} =2. $$