Suppose $b : \mathbb{R}^d\to \mathbb{R}^d$, $\sigma: \mathbb{R}^d \to \mathbb{R}^{d\times d}$ are measurable functions and satisfy $$ 2\langle x-y, b(x)-b(y)\rangle + \|\sigma(x) - \sigma(y)\|^2 \le K|x -y|\rho(|x-y|), \tag{1} $$ for all $x,y \in \mathbb{R}^d$, where $\rho: [0,\infty) \to [0, \infty)$ is some appropriate function. In addition, assume that the SDE $$ d X(t) = b (X(t))dt + \sigma(X(t))dW(t) $$ is well-posed in the strong sense.
We now consider the coupled process \begin{align*} d X(t) = b (X(t))dt + \sigma(X(t))dW(t), \\ d Y(t) = b (Y(t))dt + \sigma(Y(t))dW(t), \end{align*} where $W$ is a $d$-dimensional standard Brownian motion. Is it true that \begin{align*} &d |X(t) - Y(t)| \\ &\quad \le \rho(|X(t)- Y(t)|) dt + \left\langle \frac{X(t)-Y(t)}{|X(t)- Y(t)|}, (\sigma(X(t))-\sigma(Y(t)))dW(t)\right\rangle, \quad \forall t\ge 0? \tag{2} \end{align*}
Using Ito's formula, we have \begin{align*} d|X(t)- Y(t)|^2 &= (2\langle X(t) - Y(t), b(X(t))- b(Y(t))\rangle \\ &\qquad + \|\sigma(X(t))- \sigma(Y(t))\|^2) dt \\ &\qquad + 2\langle X(t)-Y(t), (\sigma(X(t))-\sigma(Y(t)))dW(t)\rangle. \end{align*} Using $\text{(1)}$, and apply Ito's formula to the process $|X(t)- Y(t)|^2$ using the function $f(r) = r^{1/2}$, $r\ge 0$, we can formally derive $\text{(2)}$. But there is a problem, as $f$ is not differentiable at $0$. But I don't know how to rigorously prove or disprove $\text{(2)}$.
Thanks a lot for your help.