On a Riemannian space, let $T$ be a smooth 2-tensor and let $f$ be a smooth function, with $|T|^2=f^2.$ Does it follow that $|\nabla f|^2\leq|\nabla T|^2$?
I know that $$f^2(|\nabla T|^2-|\nabla f|^2)=\frac{1}{2}|T_{ij}\nabla_kT_{pq}-T_{pq}\nabla_kT_{ij}|^2,$$ but this only gives the inequality when $f$ is nonzero.
Your computation proves the inequality at all points where $f\ne 0$. It's also clearly true at all points where $\nabla f=0$. So consider the set $S$ of points where $f=0$ and $\nabla f \ne 0$. Since you're assuming $f$ is smooth, the regular level set theorem (also known as the preimage theorem) shows that $S$ is an embedded codimension-$1$ submanifold, so its complement is dense. Therefore, by continuity, the inequality holds at points of $S$ as well.