Consider the following two statements : Statement I : For any two...

Statement I : $\left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right|$
Since \left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq\left|\frac{z_1}{\left|z_1\right|}\right|+\left|\frac{z_2}{\left|z_2\right|}\right| $$\begin{aligned} & \left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq \frac{\left|z_1\right|}{\left|z_1\right|}+\frac{\left|z_2\right|}{\left|z_2\right|} \\ & \left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2 \end{aligned}$$ \left(\left|z_1\right|+\left|z_2\right|\right)\left(\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right|\right) \leq 2\left(\left|z_1\right|+\left|z_2\right|\right)$\therefore$ statement $\mathrm{I}$ is correct For Statement II : $$\begin{aligned} & \frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|} \\ & \frac{a^2}{|y-z|^2}=\frac{b^2}{|z-x|^2}=\frac{c^2}{|x-y|^2}=\lambda \\ & a^2=\lambda\left(|y-z|^2\right)=\lambda(y-z)(\bar{y}-\bar{z}) \\ & b^2=\lambda(z-x)(\bar{z}-\bar{x}) \text{ and } c^2=\lambda(x-y)(\bar{x}-\bar{y}) \\ & \frac{a^2}{y-z}+\frac{b^2}{z-x}+\frac{c^2}{x-y}=\lambda(\bar{y}-\bar{z}+\bar{z}-\bar{x}+\bar{x}-\bar{y})=0 \end{aligned}$$ Statement II is false