I'm trying to show that given a point in the line segment connecting centers of $2$ given balls, there is a ball centered at that point inside the convex hull of the union of that $2$ other balls.
Could you have a check on my attempt? Is there other approach to this problem?
Let $(X, \|\cdot\|)$ be a normed vector space, $a,b\in X$, and $r_a , r_b \ge0$. Let $B_a$ (resp. $B_b$) be the open ball centered at $a$ (resp. $b$) with radius $r_a$ (resp. $r_b$). We fix $0 < \lambda < 1$ and let $c:=\lambda a + (1-\lambda)b$ and $r_c:=\lambda r_a + (1-\lambda)r_b$. Then $$ B_c \subset \operatorname{conv} (B_a \cup B_b). $$
My attempt: Let $x_c \in B_c$ such that $x_c \neq c$. Consider $$ x_a := a + \lambda_a (x_c - c) \quad \text{and} \quad x_b := b + \lambda_b (x_c - c) $$ for some $\lambda_a, \lambda_b >0$. We succeed if we can find $\lambda_a, \lambda_b$ such that $$ x_a \in B_a, \quad x_b \in B_b, \quad x_c = \lambda x_a + (1-\lambda)x_b, $$ or equivalently $$ \|x_a-a\| \le r_a, \quad \|x_b-b\| \le r_b, \quad x_c = \lambda x_a + (1-\lambda)x_b. $$
These conditions are equivalent to $$ \lambda_a \le \frac{r_a}{\|x_c-c\|}, \quad \lambda_b \le \frac{r_b}{\|x_b-b\|}, \quad \lambda_a\lambda+\lambda_b(1-\lambda)= 1. $$
Notice that $$ \frac{r_a}{\|x_c-c\|} \lambda+ \frac{r_b}{\|x_b-b\|} (1-\lambda) \ge \lambda + (1-\lambda)= 1. $$
So such $\lambda_a, \lambda_b$ do exist. This completes the proof.