Let $f: \mathbb R^n \to \mathbb R$ be a differentiable function such that for all $x$, $f(x) \geq f(a)$. Prove that $Df(a)$ is the zero matrix.
My solution is below, to which I request verification.
Note: This is a simple result with proofs readily available. My question is to verify this proof. In particular, did I set up the $f_i$ well to properly handle the move from single to multivariable?
For all $i \leq n$, let $f_i(t) = f(a + e_it)$, where $e_i$ is a unit vector in direction $i$.
Since $f$ is differentiable at $a$, then $f_i$ is differentiable at $0$, and $f'(a) = \sum_i f_i'(0)e_i$.
Since $f$ has a minimum at $a$, then for all $i$, $f_i$ has a minimum at $0$, and $f'_i(0) = 0$.
Therefore, $f'(a) = 0$.