prove the metric d on $\mathbb{R}^2$ defined by $d((x_1,x_2),(y_1,y_2))=\sqrt{(x_1-x_2)^2+(y_1-y_2)^4}$ satisfy the triangle inequality

Question

I'm a bit stuck with this proof. And also, what's the general strategy I need to bear in mind when proving triangle inequality?

Answer 1

We want to show

$$|d((x_{1}, x_{2}), (y_{1}, y_{2}))| \leq |d((x_{1}, x_{2}), (z_{1}, z_{2}))| + |d((z_{1}, z_{2}), (y_{1}, y_{2}))|$$

for $(x_{1}, x_{2}), (y_{1}, y_{2}), (z_{1}, z_{2}) \in \mathbb{R}^{2}.$ By the metric defined,

$$d((x_{1}, x_{2}), (z_{1}, z_{2})) + d((z_{1}, z_{2}), (y_{1}, y_{2})) = \sqrt{(x_{1} - x_{2})^{2} + (z_{1} - z_{2})^{4}} + \sqrt{(z_{1} - z_{2})^{2} + (y_{1} - y_{2})^{4}}$$

$$\geq \sqrt{(x_{1} - x_{2})^{2}} + \sqrt{(y_{1} - y_{2})^{4}}$$

$$\geq \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{4}} = d((x_{1}, x_{2}), (y_{1}, y_{2})). $$

The last inequality follows from this proof here.