My task is to show the following: Given $x \in \mathbb{R}^d$ and let $v_d$ be a constant represent the volume of the unit hypersphere in $\mathbb{R}^d$, show that $$\int_{\mathbb{R}^d} 1_{\{ \lVert x + b \rVert < 1 \le \lVert b \rVert\}}db \le \min\left\{v_d, \frac{\pi}{2} v_{d - 1} \lVert x \rVert\right\}$$ where $\lVert{}\cdot{}\rVert$ is the usual Euclidean norm on $\mathbb{R}^d$.
This integral can be thought as the volume of the unit ball centered at $-x$ that is outside the unit ball centered at the origin. Little bit of algebra can transformed the integral into: $$\int_{\mathbb{R}^d} 1_{\lVert x + b \rVert < 1, \lVert b \rVert < 1}db \ge \max\left\{0, v_d - \frac{\pi}{2} \lVert x \rVert v_{d - 1}\right\}$$
What I have done is to first prove this in the case when $d = 2$. Then, in this case $v_2 = \pi$ and $v_1 = 2$, and the integral becomes: $$\int_{\mathbb{R}^2} 1_{\{\lVert x + b \rVert < 1, \lVert b \rVert < 1\}}db \ge \max\left\{0, \pi(1 - \lVert x \rVert)\right\}$$
Obviously, when $\lVert x \rVert \ge 1$ or $x = 0$, $\displaystyle\int_{\mathbb{R}^2} 1_{\{\lVert x + b \rVert < 1, \lVert b \rVert < 1\}}db \ge 0$. Now, consider the case $0 < \lVert x \rVert < 1$, and change the integral into polar coordinate, we need to show that $$\int_0^1 \int_0^{2\pi} 1_{\{\vert -x_1 + r\cos \varphi \vert^2 + \vert -x_2 + r\sin \varphi \vert^2 < 1\}}rd\varphi dr \ge \pi(1 - \lVert x \rVert)$$
If we rewrite $x$ in the polar coordinates: $x = (\widetilde{r}\cos \widetilde{\varphi}, \widetilde{r}\sin \widetilde{\varphi})$ with $\widetilde{r} \in (0, 1)$, $\widetilde{\varphi} \in [0, 2\pi)$, then it becomes: $$\int_0^1 \int_0^{2\pi} 1_{\{r^2 + \widetilde{r}^2 - 2r\widetilde{r}\cos(\varphi - \widetilde{\varphi}) < 1\}}rd\varphi dr \ge \pi(1 - \widetilde{r})$$
Here is where I got stuck and I don't know how to progress this. So, if you have any idea to progress this, I'd love to hear. Thank you.
Update 1: After tinkering a bit in R code, I find that the circle centered at the midpoint $-x/2$ with radius $\sqrt{1 - \lVert x \rVert}$ lies perfectly in the intersection:
Here's the code I use to generate this image:
r = runif(1, 0, 1)
phi = runif(1, 0, 2*pi)
x1 = r*cos(phi)
x2 = r*sin(phi)
b1 = c()
b2 = c()
for(i in 1:10^5){
r = runif(1, 0, 1)
phi = runif(1, 0, 2*pi)
if((x1 + r*cos(phi))^2 + (x2 + r*sin(phi))^2 < 1)){
b1 <- c(b1, r*cos(phi))
b2 <- c(b2, r*sin(phi))}
}
flag = (b1 + x1/2)^2 + (b2 + x2/2)^2 < 1 - sqrt(x1^2 + x2^2)
plot(b1[!flag], b2[!flag], col = 'red')
points(b1[flag], b2[flag], col = 'blue')
So now, it is only sufficient to show that for $b \in \mathbb{R}^2$, $\left\lVert b + \dfrac{x}{2}\right\rVert < \sqrt{1 - \lVert x \rVert}$, then we must have $$\begin{cases} \lVert b \rVert < 1\\ \lVert x + b \rVert < 1 \end{cases}$$
Update 2: I'll start a bounty on this question since I can't generalized my answer below from $\mathbb{R}^2$ into $\mathbb{R}^d$.
$\def\d{\mathrm{d}}\def\R{\mathbb{R}}\def\y{\tilde{y}}\def\ge{\geqslant}\def\le{\leqslant}\def\Ω{{\mitΩ}}\def\peq{\mathrel{\phantom{=}}}\def\T{^{\mathrm{T}}}$This answer proves that:
(Note that $\dfrac{\pi}{2} > 1$.)
Proof: It's trivial that $\displaystyle\int_{ \substack{ \|y\| > 1 \\ \|y - x\| \le 1 } } \d y \le v_d$.
Now since it's the Euclidean norm, it can be assumed without loss of generality that $x = (x_1, 0, \cdots, 0)\T$, where $x_1 \ge 0$. (See Appendix.) Thus $\|x\| = x_1$. For any $y \in \R^d$, denote by $y_1$ the first coordinate of $y$, and by $\y$ the point in $\R^{d - 1}$ formed by the last $d - 1$ coordinates of $y$.
For any $y \in \R^d$,\begin{align*} &\mathrel{\phantom{\implies}}\begin{cases} \|y\| > 1\\ \|y - x\| \le 1 \end{cases} \iff \begin{cases} y_1^2 + \|\y\|^2 > 1\\ (y_1 - x_1)^2 + \|\y\|^2 \le 1 \end{cases}\\ &\implies \begin{cases} \|\y\| \le 1\\ y_1 \in (-\infty, -\sqrt{ 1 - \|\y\|^2 }) \cup (\sqrt{ 1 - \|\y\|^2 }, +\infty)\\ y_1 \in [x_1 - \sqrt{ 1 - \|\y\|^2 }, x_1 + \sqrt{ 1 - \|\y\|^2 }] \end{cases}\\ &\stackrel{ x_1 \ge 0 }{\implies} \begin{cases} \|\y\| \le 1\\ \sqrt{ 1 - \|\y\|^2 } \le y_1 \le x_1 + \sqrt{ 1 - \|\y\|^2 }, \end{cases} \end{align*} therefore\begin{gather*} \int\limits_{ \substack{ \|y\| > 1 \\ \|y - x\| \le 1 } } \d y \le \iint\limits_{ \substack{ \|\y\| \le 1 \\ \sqrt{ 1 - \|\y\|^2 } \le y_1 \le x_1 + \sqrt{ 1 - \|\y\|^2 } } } \d y_1 \d \y\\ = \int\limits_{ \|\y\| \le 1 } \d \y \int_{ \sqrt{ 1 - \|\y\|^2 } }^{ x_1 + \sqrt{ 1 - \|\y\|^2 } } \d y_1 = x_1 \int\limits_{ \|\y\| \le 1 } \d \y = x_1 v_{d - 1} = \|x\| v_{d - 1}. \tag*{$\Box$} \end{gather*}
Appendix: For $x \ne 0$, augment it into a basis $\{x, u_2, \cdots, u_d\}$ for $\R^d$, and the Gram-Schmidt process yields an orthonormal basis $\{v_1, \cdots, v_d\}$, where $v_1 = \dfrac{x}{\|x\|}$.
Define $\Ω = (v_1, \cdots, v_n)$, $φ(u) = \Ω u$ ($u \in \R^d$), then $φ$ is an orthogonal transformation. Note that$$ φ((\|x\|, 0, \cdots, 0)\T) = \|x\| v_1 = x \implies φ^{-1}(x) = (\|x\|, 0, \cdots, 0)\T. $$ Denote $D_y = \{y \in \R^d \mid \|y\| > 1,\ \|y - x\| \le 1\}$, then\begin{align*} φ^{-1}(D_y) &= \{u \in \R^d \mid \|φ(u)\| > 1,\ \|φ(u) - x\| \le 1\}\\ &= \{u \in \R^d \mid \|u\| > 1,\ \|u - φ^{-1}(x)\| \le 1\} =: D_u, \end{align*} and the change of variable $y = φ(u)$ yields\begin{gather*} \int_{D_y} \d y = \int_{D_u} |\det φ'(u)| \,\d u = \int_{D_u} |\det \Ω| \,\d u = \int_{D_u} \,\d u.\tag*{$\Box$} \end{gather*} Remark: Technically speaking, only if $\det \Ω = 1$ is $φ$ a rotation, but for this answer it suffices to have $\det \Ω = \pm 1$.