Suppose $B_t$ is a standard Brownian motion on a filtered probability space $\langle \Omega, \mathcal F, \{\mathcal F_t\}_t, \mathbb P\rangle$. Consider two SDEs below.

Suppose, $X_0 = Y_0 = 0$ \begin{align*} dX_t =& sign(X_t) dt + d B_t\\ dY_t =& \alpha_t dt + dB_t \end{align*} where $\alpha_t$ is some $[-1,1]$-valued, $\mathcal F_t$ measurable process.

I am wondering if the following is true. And if yes, how might one approach this thing? $$\mathbb E[\vert X_T\vert] \ge \mathbb E[\vert Y_T \vert]$$ for all $T \ge 0$.

Intuitively, the reason for this conjecture is the following: Both the processes start from $0$. Say, at some time they are equal to some $x > 0$. Now, $X_t$ has an upward drift of $1$ while the upward drift of $Y_t$ is weakly less than $1$. So, the moment the two processes have the same sign, it seems that $X_t$ runs farther away compared to $Y_t$. But, I have no idea how to make this precise, even if true.

Thanks!