Suppose $\Omega \subset \mathbb{R}^3$ is a bounded smooth domain. Let $f: \mathbb{R} \to \mathbb{R}$ and $u : \Omega \to \mathbb{R}$ be smooth functions. I would like to prove $$ \int_\Omega f(u(x))|\nabla u| dx \geq \int_{\mathbb{R}} f(y) \mathcal{H}^2\Big(\Omega \cap \{x\in \Omega : u(x)=y\}\Big) dy.$$ By the co-area formula, I can only get $$ \int_\Omega f(u(x))|\nabla u| dx \geq \int_{\mathbb{R}} \left(\int_{u^{-1}(y)} f\big(u(z)\big) d\mathcal{H}^2_z\right)dy.$$ It seems like the result is straightforward if we replace $z$ by $u^{-1}(w)$ intuitively. But how can we do that?
I have tried to approximate $f\big(u(x)\big)$ by simple function so that $f$ can be treated as a constant and take it from the integral, but i have some troubles in this way.
I don't know if I'm missing something, but I think that this follows from $z \in u^{-1}(y)$ if and only if $u(z)=y$. Therefore, $$ \int_{u^{-1}(y)} f(u(z))d\mathcal H_{z}^{2}=\int_{u^{-1}(y)} f(y)d\mathcal H_{z}^{2}=f(y)\mathcal{H}^{2}(u^{-1}(y)).$$