Denote the center of six circles by $O_i$ and their radii by $r_i$ where $i\in \{ 1,2,...,6\}$. Let $O$ and $r$ be the center and radius of the circle intersecting all six circles. Consider angle formed by rays $\overrightarrow{OO_i}$ where $i\in \{ 1,2,...,6\}$. Suppose that $r<1$. Not hard to show that there exist $i,j\in \{ 1,2,...,6\}$ that $i\neq j$ and $\angle{O_iOO_j}\leq 60^{\circ}$. Law of cosine gives us $O_iO_j^2=OO_i^2+OO_j^2-2\times OO_i\times OO_j\times \cos( \angle{O_iOO_j})\leq OO_i^2+OO_j^2-OO_i\times OO_j$. WLOG $OO_i\geq OO_j$, we get $OO_i\times OO_j\geq OO_j^2\Rightarrow 0\geq OO_j^2-OO_i\times OO_j$. But we also have $O_iO_j>r_i+r_j\geq r_i+1>r_i+r\geq OO_i$, so $O_iO_j^2> OO_i^2$. Hence $O_iO_j^2> OO_i^2+OO_j^2-OO_i\times OO_j$, contradiction. So $r\geq 1$, done.

Same notation as ThE-dArK-lOrD used, but more synthetic argument: After showing that there exists such $\angle O_iOO_j \le 60^\circ$ we could say that then (WLOG) $r+r_i \ge OO_i \ge O_iO_j > r_i+r_j$, which implies $r > r_j \ge 1\square$

Isn't this problem flavour/main argument suspiciously similar to IMO Shortlist 2020 G4? (EDIT: I just saw this has been mentioned in the early posts in the IMO SL thread)