I am reading a proof for the existence of a solution to the Local Cauchy problem to the non-linear Schrodinger equation $$ i\partial_t u+\Delta u +\epsilon u |u|^{2} = 0 \\ u(x,0)=u_0(x) $$

The structure of the proof is due to J. Ginibre and G. Velo (On a class of nonlinear Schrödinger equations. I. The Cauchy problem, general case). Details of the proof are NOT necessary for understanding what I am asking here

Denote, by $L_{T}^p L_x^q$ the spacetime banach space $L([0,T]; L^q (\mathbb R^2))$. The norm is given by $$ \left(\int_0^T \|u(\cdot, t)\|_{L^q (\mathbb R^2)}^p dt\right)^{1/p}. $$

One line in the proof seems to say that for all functions $u, v$ with sufficient regularity (say, in $C_c^\infty(\mathbb R^d \times \mathbb R)$), we have for some $C>0$ independent of $u,v$, $$ \|\nabla(u|u|^2)-\nabla(v|v|^2)\|_{L^1_TL^2_x}\\ \leq C(\||\nabla(u-v)| (|u|^2+|v|)\|_{L^1_TL^2_x} \\ + \| |u-v|(|u|+|v|)(|\nabla u|+|\nabla v|) \|_{L^1_TL^2_x}). $$

I do not understand why this inequality holds. How to prove this inequality?

I proved a similar inequality regarding $u$ rather than $\nabla u$ by showing that for complex numbers $u,v$, $|u|u|^2-v|v|^2|\leq C |u-v|(|u^2|+|v^2|)$, but it seems to be really hard to extend this to $\nabla(u|u|^2)$.