Kato's Inequality with Laplacian Let $u \in C^2(\mathbb R^n)$ be complex-valued. I am trying to prove the inequality 
$$\Delta |u| \geq \Re\left(\frac{\bar u}{u} \Delta u\right),$$ 
in the distributional sense. I tried calculating both sides explicitly in terms of the real and imaginary parts of $u$, but the expressions I get don't seem to be directly comparable. Is there a simple proof?
 A: There is almost certainly a beautiful proof that generalizes this inequality to elliptic operators and makes use of some nice, general properties, but here is an elementary proof using direct calculation: Set $|u| = (u \bar{u})^{1/2}$, and calculate componentwise:
$\begin{align*}
\Delta |u| & = \sum_i \frac{1}{2} |u|^{-1} D^2_{x_i}|u| - \frac{1}{4} |u|^{-3} [D_{x_i} |u|]^2 \\
& = \sum_i \frac{1}{2} |u|^{-1} (\bar{u} u_{x_i x_i} + 2 u_{x_i} \bar{u}_{x_i} + u \bar{u}_{x_i x_i}) - \frac{1}{4} |u|^{-3} (u^2 \bar{u}_{x_i}^2 + 2|u|^2 u_{x_i} \bar{u}_{x_i} + \bar{u}^2 u_{x_i}^2 ) \\
& = \frac{1}{2} |u|^{-1} (\bar{u} \Delta u + u \Delta \bar{u}) - \frac{1}{4} |u|^{-3} \sum_i (u \bar{u}_{x_i} - u_{x_i} \bar{u})^2
\end{align*}$
Now we claim that each term $u \bar{u}_{x_i} - u_{x_i} \bar{u}$ is purely imaginary. One can check this easily (write $u = f + ig$ with $f,g$ purely real). Therefore the square of such a term is nonpositive, so in the above inequality we are subtracting a nonpositive term. Thus we have
$\begin{align*}
\Delta |u| & \geq \frac{1}{2} |u|^{-1} (\bar{u} \Delta u + u \Delta \bar{u}) \\
& = \Re\left[ \frac{\bar{u}}{|u|} \Delta u \right].
\end{align*}
$
Edit: This differs from the inequality you stated above, but as far as I can tell, this must be the correct version of the inequality, because if $u$ is purely real, then a direct calculation of $\Delta |u|$ yields
$$\Delta |u| = \frac{u}{|u|} \Delta u$$
whereas in the inequality you stated, were it true, then for $u$ purely real, we would have $\Delta |u| \geq \Delta u$, which is false if $u(x) < 0$ but $\Delta u(x) > 0$.
