Normalized spaces. Prove that if one closed ball nested into other one, then $r_1 \leq r_2 - ||x_1-x_2||$ Let $(X, ||*||)$  normalized spase.
Prove that if one closed ball nested into other one ($\overline{B_{r_1}}(x_1) \subset \overline{B_{r_2}}(x_2))$, then $r_1 \leq r_2 -||x_1-x_2||$, where $r_1, r_2$ are radiuses of such balls and $x_1, x_2$ are their centers.
I tried to transform this inequality into: $||x - y|| \geq r_2 - r_1$, and make something with it. Because we know that: $||x - y|| \geq ||x|| - ||y||$.
 A: If $x_1 = x_2$, then the result is obviously true. So let's suppose that $x_1 \neq x_2$. Consider
$$x'' = x_1 + \frac{r_1}{||x_1-x_2||}(x_1-x_2)$$
You have
$$||x''-x_1|| = \frac{r_1}{||x_1-x_2||}||x_1-x_2||=r_1$$
so $x'' \in \overline{B}(x_1, r_1)$. Because $\overline{B}(x_1, r_1) \subset \overline{B}(x_2, r_2)$, you deduce that $x'' \in \overline{B}(x_2, r_2)$, i.e.
$$||x''-x_2|| \leq r_2 \quad \quad (*)$$
But $$||x''-x_2|| = \left|\left|x_1 + \frac{r_1}{||x_1-x_2||}(x_1-x_2)-x_2\right|\right|$$
$$=\left|\left|\left(x_1-x_2 \right) \left( 1 + \frac{r_1}{||x_1-x_2||}\right)\right|\right|$$
$$= \left( 1 + \frac{r_1}{||x_1-x_2||}\right) ||x_1-x_2|| = ||x_1-x_2|| + r_1$$
So the relation $(*)$ gives directly $$||x_1-x_2|| + r_1 \leq r_2$$
which is exactly what you asked for.
A: Careful, you want to prove $\|x_1-x_2\| \le r_2-r_1$, not $\,\ge$.
Assume $x_1\ne x_2$ and define $y = (1-\lambda) x_1+\lambda x_2$.
If we set $\lambda = \frac{r_1}{\|x_1-x_2\|}$ then $$\|x_1-y\| = \lambda\|x_1-x_2\| = r_1$$ so $y \in \overline{B}(x_1,r_1)$. Therefore $y \in \overline{B}(x_2,r_2)$ as well so
$$\|x_1-x_2\| - r_1 = (1-\lambda)\|x_1-x_2\| = \|y-x_2\| \le r_2$$
which proves the desired claim.
