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, $f(y)$ > $f(a)$.
 A: You have $f(a)<f(x)+\epsilon$ and $f(x)<f(y).$ This does NOT imply $f(a)<f(y). $
(1). Suppose $c\in (b,a)$ and $f(c)< f(a).$
Let $e= (f(a)-f(c))/2.$ Let  $d\in (c,a)$ such that  $f(d)-f(a)>-e.$ We know that $d$ exists because $f$ is continuous at $a.$ By the MVT there exists $d'\in (c,d)$ such that $$f'(d')=\frac {f(d)-f(c)}{d-c}>\frac {f(a)-e-f(c)}{d-c}=\frac {e}{d-c}>0$$  contrary to $\forall d'\in (b,a)\;(f(d')<0).$
Therefore $\forall c\in (b,a)\;(f(c)\ge f(a)\,).$
(2). Suppose $c'\in (b,a)$ and $f(c')=f(a).$ Let $c=(a+c')/2.$ By (1) we have $f(c)\ge f(a).$ By the MVT there exists $c''\in (c,c')$ such that $$f'(c'')=\frac {f(c)-f(c')}{c-c'}\ge \frac {f(a)-f(c')}{c-c'}=\frac {f(c')-f(c')}{c-c'}=0$$ contrary to $\forall c''\in (b,a)\;(f'(c'')<0).$
Therefore $\forall c'\in (b,a)\;(f(c')\ne f(a)\,).$
(3). By (1) and (2) we have $\forall x\in (b,a)\;(f(x)>f(a)\,).$
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. 
