Proving that the intersection of $(a-4)^{2}+(b-1)^{2}<4$ and $(a-1)^{2}+(b-5)^{2}<4$ is empty I´m trying to prove that these two inequalities intersection is empty, but i don´t know how to proceed i know they are circles, but i´m trying to do it with only algebra
Let $a,b\in\mathbb{R}$ and the inequalities are
$$(a-4)^{2}+(b-1)^{2}<4 \quad\land\quad (a-1)^{2}+(b-5)^{2}<4$$
Any insight would be appreciated.

Lets suppose that $(a,b)$ belongs to the set of points where
$$ \sqrt{(a-1)^2+(b-5)^2} <2. $$
By the triangle inequality for vectors in $\mathbb R^2$, in the form $|x|\ge |x-y|-|y| $, where $|x|:=\sqrt{x_1^2+x_2^2}$,
$$\sqrt{(a-4)^2+(b-1)^2} \ge \sqrt{(4-1)^2+(1-5)^2}-\sqrt{(a-1)^2+(b-5)^2}   > 5 - 2 =3 $$
which implies that $(a,b)$ cannot belong to the set of points where
$$ \sqrt{(a-4)^2+(b-1)^2} < 2.$$ So there are no points in the intersection.
