Proving one open ball is a subset of another open ball

Question

This is a smaller part of a larger proof but I just can't get past it. I need to prove $B_r(y) \subset B_4(x)$ for $x$ and $y$ being the points $(0,0)$ and $(1,3)$ respectively and the radius $r$ being $(4-\sqrt{10}).$ All I've come up with so far is that by expanding the formula for the circle around $B_r(y)$ I need to show $x^2+y^2<2x+6y+16-8 \sqrt{10}$ implies $x^2+y^2<16$. I'm just unsure of how to relate the two. I've also attempted to use the contrapositive and met a similar problem.

Answer 1

Assuming the usual Euclidean metric in $\mathbb{R}^2$, $d(x,y)=\sqrt{10}$. Then by the triangle inequality, for any $z$, $$ d(x,z)\leq d(x,y)+d(y,z)=\sqrt{10}+d(y,z). $$

So if $d(y,z)\leq 4-\sqrt{10}$, the claim follows.