Is this vector calculus version of Lagrange's identity correct?

Question

Lagrange's identity can be written like $$ |\mathbf{a}|^2|\mathbf{b}|^2 = |\mathbf{a}\times \mathbf{b}|^2 + |\mathbf{a}\cdot \mathbf{b}|^2 $$ Now if we replace the vector $\mathbf{a}$ with the Nabla operator $\nabla$ we get $$ |\nabla|^2|\mathbf{b}|^2 = |\nabla\times \mathbf{b}|^2 + |\nabla\cdot \mathbf{b}|^2 $$ or using the Laplace operator $\Delta$ $$ \Delta (\mathbf{v} \cdot \mathbf{v}) = |\nabla\times \mathbf{v}|^2 + |\nabla\cdot \mathbf{v}|^2 $$ This looks nice, but is it true? Can one derive and prove this identity directly via vector calculus (or by any other means)?

Answer 1

Consider this example

$$ \mathbf{v} = \frac{1}{2}x^2 \hat{x} $$

In this case

$$ \nabla\cdot \mathbf{v} = x $$

and

$$ \nabla \times \mathbf{v} = 0 $$

Therefore

$$ |\nabla \cdot \mathbf{v}|^2 + |\nabla \times \mathbf{v}|^2 = x^2 \tag{1} $$

On the other hand

$$ \Delta (\mathbf{v}\cdot \mathbf{v}) = \Delta(x^4/4) = 3x^2 \tag{2} $$

From (1) and (2):

$$ \Delta (\mathbf{v}\cdot \mathbf{v}) \ne |\nabla \cdot \mathbf{v}|^2 + |\nabla \times \mathbf{v}|^2 $$