[Math] Triangle inequality and the square root of a metric space

real-analysis

enter image description here

for (i) I know that the square root part is true but I don't know how to put it into words to prove it.

For (ii) I just don't know how top apply the requirements for a metric space to the square root of another metric space. Just kind of confusing me

Best Answer

Suppose to the contrary that $\sqrt{a}\gt \sqrt{b}+\sqrt{c}$. Then $$a\gt (\sqrt{b}+\sqrt{c})^2=b+c+2\sqrt{bc}\ge b+c,$$ so $a\gt b+c$, contradicting the fact that $a\le b+c$.

Best Answer

Related Solutions

Proving triangle inequality for: $(X,d)$ is a metric space. Let $d′(x,y) = \min \{1, d(x,y)\}$.Prove that $(X,d′)$ is also a metric space.

Triangle Inequality for $\ell^2$ Metric

Related Question