Inequality for the expected values of norm of stochastic processes Let $\underline{X}=(x_1, x_2, x_3), \; x_i \sim \mathcal{N(0,1)}$ i.i.d.
For any fixed $t>0$ and $\underline{X}_0$ prove that the following holds ($\Vert\cdot\Vert$ is the Euclidean norm):
$$\mathbb{E}(\Vert \underline{X}_0+\sqrt{t}\underline{X} \Vert^{-1}) \leq \mathbb{E}(\Vert \sqrt{t}\underline{X} \Vert^{-1}) $$

It is quite easy to convince yourself with the following argument:
$\underline{B}_t \sim \underline{X}_0+\sqrt{t}\underline{X}$ (where $\underline{B}_t$ denotes Brownian motion)
Then the desired result is equivalent to this:
$ \mathbb{E}_{\underline{X}_0}(\Vert \underline{B}_t \Vert^{-1}) \leq \mathbb{E}_{0}(\Vert \underline{B}_t \Vert^{-1})$ (where $\mathbb{E}_{\underline{X}_0}$ denotes that the Brownian motion starts at $\underline{X}_0$)
And intuitively it seems to be quite obvious.
On the other hand, I can't find a short proof of it. Thank you very much in advance!
 A: 
Key-facts: The function $x\mapsto\|x\|^{-1}$ is harmonic on $\mathbb R^3\setminus\{0\}$ and the distribution of $\sqrt{t}\underline{X}$ is radially symmetric.

Assume that $\underline{X}_0$ and $\underline{X}$ are independent (otherwise anything can happen) and note that the result is a consequence of the fact that, for every random variable $B$ whose distribution is radially symmetric in $\mathbb R^3$ and for every $x$ in $\mathbb R^3$, 

$$m(x)\leqslant m(0),\qquad m(x)=E(h(x+B)),\qquad h:x\mapsto\|x\|^{-1}.$$ 

To prove this, first note that, for every rotation $r$, $r^{-1}(B)$ and $B$ are identically distributed and $h\circ r=h$ hence $$m(x)=E(h(x+r^{-1}(B)))=E(h(r(x)+B))=m(r(x)).$$ Averaging this identity over every rotation and considering some random variable $U_x$ uniform on the sphere of radius $\|x\|$ and independent of $B$, one gets $m(x)=E(m(U_x))$, that is, using the independence, $$m(x)=E(h(U_x+B))=E(u_x(B)),\qquad u_x(y)=E(h(U_x+y)).$$
The function $h$ is harmonic on $D=\mathbb R^3\setminus\{0\}$ hence its average of on every sphere bounding a ball included in $D$ equals its value at the center of the sphere, that is, for every $\|y\|\geqslant\|x\|$, $$u_x(y)=h(y).$$ On the other hand, $u_x$ is harmonic on the ball $\|y\|\leqslant\|x\|$ and $u_x=h(x)$ on the boundary hence, for every $\|y\|\leqslant\|x\|$, $$u_x(y)=h(x).$$
To sum up, $u_x=\min(h(x),h)$ hence $$m(x)=E(\min(h(x),h(B)))\leqslant E(h(B))=m(0).$$
