Let $\Omega\subset\mathbb R^n$ be open and connected and let $f\in W^{1,2}(\Omega)$. Then Korn's inequality tells us that there exists $C>0$ such that $$\|f\|_{W^{1,2}}^2\leq C(\|f\|_{L^2}^2+\|\mathrm{Sym}\, df\|_{L^2}^2), $$ where $(\mathrm{Sym}\, df)_{ij}=\frac{1}{2}(\partial_i f^j+\partial_j f^i)$.
How does it follow that there exists a (constant) skew-symmetric matrix $w$ (which may depend of $f$) such that $$\|df-w\|_{L^2}\leq C'\|\mathrm{Sym}\, df\|_{L^2}?$$
I thought about decomposing $f$ into its symmetric and skew-symmetric parts, namely $f=\mathrm{Sym}\, df+w$, where $w=\frac{1}{2}(df-df^T)$. But then I realized that $w$ has to be a constant matrix, so that doesn't really help.