estimate on $| \nabla (u |u|^2) - \nabla(w|w|^2)|$ for $u,w \in H^1$ suppose $u, w \in H^1 (R^2)$. I'd like to know where does the following inequality come from (it appears in a proof I've been reading and I can't figure it out)
$$ | \nabla (u |u|^2) - \nabla(w|w|^2)| \leq C | \nabla(u-w)| \cdot(|u|^2 + |w|^2)
 + C|u-w|\cdot(|\nabla u| + |\nabla w|)(|u|+|w|)$$
$u$ and $w$ are complex valued. Can anyone provide some suggestions? C is some constant, its value isn't importnat.
Perhaps I should mention that I'm going to use it with integrals ($L^2$), but I'm pretty sure it should be a pointwise inequality
 A: Let $f:\mathbb{C}\to\mathbb{C}$ be the function $f(x,y) = (x+iy) (x^2 + y^2)$. Its derivative is
$$ \mathrm{d}f = (x^2 + y^2) \mathrm{d}(x + i y) + 2(x + i y) (x \mathrm{d}x + y \mathrm{d}y) $$
which satisfies
$$ |\mathrm{d}f| \leq C(x^2 + y^2) $$
Integrating along the straight line joining $v,w\in \mathbb{C}$ gives
$$ |f(w) - f(v) | \leq C \max(|v|^2,|w|^2) |v-w| \leq C(|v|^2 + |w|^2) |v-w|$$ 
Now let $w,v$ be differentiable $\mathbb{C}$ valued functions. 
Since you are applying to functions in $L^2$ this is okay by density. 

Since there seems to be some confusion, let me do this in a bit more detail. 
Let $u_s = (1-s) v + s w$ so that $u_0 = v$ and $u_1 = w$ be a one parameter family of functions. What we should do is to compute
$$ \nabla f(v) - \nabla f(w) = \nabla f(u_0) - \nabla f(u_1) = \int_1^0 \partial_s \nabla f(u_s) \mathrm{d}s $$
The term (writing $u = u_s$ and $\dot{u} = \partial_s u_s$)
$$ \partial_s \nabla f(u) = \nabla \left[ 2\bar{u} u \dot{u} + u^2 \dot{\bar{u}}\right] $$
Putting in the $\nabla$ you see that by the product rule you have
$$ \partial_s \nabla f(u) = 2 \nabla \bar{u} u \dot{u} + 2 \bar{u} \nabla u \dot{u} + 2 \bar{u} u \nabla \dot{u} + 2 u \nabla u \dot{\bar{u}} + u^2 \nabla \dot{\bar{u}} $$
Now, since $u$ is linear in $s$, for each component of the $\nabla$ the term with $\dot{u}$ is signed. So putting in absolute value signs and integrating both sides give you the desired inequality. 
