If $f$ is continuous at $a$ and $f' < 0$ on $(b, a)$, then $a$ is a minimum on $(b, a)$. Could you please verify my proof? I feel like it's terribly overwrought and I'm missing a much simpler explanation. (It may even be invalid, see the bold).

If $f$ is continuous at $a$ and $f' < 0$ on $(b, a)$, then $a$ is a minimum on $(b, a)$.
$\forall y, x \ \ \ \ \ \ \ \ \ a-\delta < y < x < a \Rightarrow f(a) < f(x) + \epsilon$
Since $y, x$ are in $(a-\delta, a)$, then $y,x $ are in $(b, a)$. Hence $f(y) > f(x)$, and $f(y) - f(x) > 0$.
Therefore,
$\forall y, x \ \ \ \ \ \ \ \ \ a-\delta < y < x < a \Rightarrow f(a) < f(y)$
Showing that $a$ is a minimum on $(a - \delta,a)$, and consequently $(b,a)$, since for any $y$ < $a$ in the former interval, $f(y)$ > $f(a)$.

If $f'<0$ then for any $x<y$, by the mean value theorem, there's a $\xi\in(x, y) $ such that $f(y)-f(x)=f'(\xi)(y-x) <0 $, hence $f(y)<f(x)$, i.e. $f$ is strictly decreasing. 
