Another way to prove $d(x,y)$ is a metric

by DavidS   Last Updated October 16, 2018 21:20 PM - source

Today on an exam, I was tasked with proving that for all $$x,y$$ in $$R_+^n$$, $$d(x,y)=\sqrt{\sum_{i=1}^n(\sqrt{x_i}-\sqrt{y_i})^2 }$$ is a metric. It's pretty clear that $$d(x,y)=0$$ iff $$x=y$$, and also that $$d(x,y)=d(y,x)$$, but when it came to verifying the triangle inequality, I did not know what to do.

In the end, I resorted to the fact that if $$d(x,y)$$ is a metric and $$f(x)$$ is an injective function over the domain of $$x$$, then $$d(f(x),f(y))$$ is a metric as well. In this case, the metric I was told to prove is similar in form to the Euclidean metric, where $$f(x)=\sqrt{x}$$ is the injective function.

This may seem totally fine, except for the fact that we were never taught this theorem in class, so I doubt that we were expected to use it.

Can someone show me another way to prove that $$d(x,y)$$ is a metric other than using the aforementioned theorem? If not, a proof of the theorem I used would also be greatly appreciated.

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You can do it as follows:\begin{align}d\bigl((x_1,\ldots,x_n),(z_1,\ldots,z_n)\bigr)&=\sqrt{\sum_{k=1}^n\left(\sqrt{x_k}-\sqrt{z_k}\right)^2}\\&=\left\|\left(\sqrt{x_1},\ldots,\sqrt{x_n}\right)-\left(\sqrt{z_1},\ldots,\sqrt{z_n}\right)\right\|_2\\&\leqslant\left\|\left(\sqrt{x_1},\ldots,\sqrt{x_n}\right)-\left(\sqrt{y_1},\ldots,\sqrt{y_n}\right)\right\|_2+\\&\phantom{=}+\left\|\left(\sqrt{y_1},\ldots,\sqrt{y_n}\right)-\left(\sqrt{z_1},\ldots,\sqrt{z_n}\right)\right\|_2\\&=\sqrt{\sum_{k=1}^n\left(\sqrt{x_k}-\sqrt{y_k}\right)^2}+\sqrt{\sum_{k=1}^n\left(\sqrt{y_k}-\sqrt{z_k}\right)^2}.\end{align}