If $f$ is continuous on $[a,\infty)$ and increasing on $(a,\infty)$, can we say $f(x)>f(a)$. If $f$ is continuous on $[a,\infty)$ and increasing on $(a,\infty)$ can we say $f(x)>f(a)$ for $x>a$. 
I think yes, the fact that $f$ is (strictly) increasing on $(a,\infty)$ gives,
$$f(x)>f(y)$$
For $x,y \in (a,\infty)$ with $x>y$, in particular take $y=a+.0000000....01$ or $y=a+\epsilon$ with $\epsilon>0$, the inequality holds and $f(a+\epsilon) \to f(a)$ by continuity. 
But I don't know how to make a rigorous proof.
 A: Let $x>a$, let $t$ be in $(a,x)$, and assume that $f(a)>f(t)$.  
Since $f$ is continuous on $[a,t]$, by the Intermediate Value Theorem 
there is a number $c$ in $(a,t)$ with $\displaystyle f(c)=\frac{f(a)+f(t)}{2}>\frac{f(t)+f(t)}{2}=f(t)$.  
Since $c<t$ with $f(c)>f(t)$,  this gives a contradiction.
Therefore $f(a)\le f(t)<f(x)$.

(edited to correct the error pointed out by Furrane)
A: This is true, the problem is only weak inequalities are preserved under limits (can you prove it?). You can overcome this: Take $z>a$, then there are $x,y$ such that $z>y>x>a$ (why?). We have $f(z)>f(y)>f(x)$. Take $x\rightarrow a$.
A: This holds true if $f$ is strictly increasing, which means :
$$\forall x \in (a,+\infty), \forall y>x : f(y)>f(x)$$
In particular :
$$\forall x\in (a,+\infty), f(x)>f\left(a+{x-a\over2}\right)\ge f(a)$$
But if $f$ in increasing (not strictly) well you could have for example :
$$\forall x \in (a,+\infty), f(x)=f(a)$$
Which is obviously not compatible with $f(x)>f(a)$
If $f$ is increasing, the best comparaison you can have is :
$$\forall x \in (a,+\infty), \forall y>x : f(y)\ge f(x)$$
Which would lead to :
$$\forall x \in (a,+\infty), f(x)\ge f(a)$$
A: Correct me if wrong:
Let $f$ be continuous on $[a,n]$, $n \in \mathbb{N}$, and strictly increasing on $(a,n]$.
1) min {$f(x) | x \in [a,n]$} = $f(a)$.
The continuous function $f$ on a closed interval attains its minimum.
Assume the function has its minimum at $x_0 \gt a$.
Then $f((x_0 + a)/2)  \lt f(x_0)$ since $f$ is strictly increasing on $(a,n]$.
Contradiction, since $f(x_0)$ was assumed to be the minimum.
Hence min {$f(x)| x \in [a,n]$} = $f(a)$.
2) $f(a) \lt f(x)$ for $x \gt a$.
Since $f(a)$ is the minimum on $[a,n]$ we have
$f(a) \le f(x)$ , $x \in [a,n]$.
It remains to be  shown  that 
$f(a) \lt f(x)$ , $x \in (a,n]$.
Assume for an $x_1 \gt a$ we have $f(a) = f(x_1)$.
Consider $x_2 = (x_1 + a)/2  \lt x_1$.
It follows $f(x_2) \lt  f(x_1) = f(a)$, the minimum, since $f$ is stricly increasing and  $x_2  \lt x_1$.
A contradiction. 
Now the other case:
Let $f$ be increasing in $(a,n]$, $n \in \mathbb{N}$.
1) $f$ has its minimum at $x = a$ as before, 
I.e. $f(a) \le f(x)$ for $x \in (a,n]$.
Proof by contradiction:
Assume there is an $x_1 \gt a$ such that $f(x_1) \lt f(a)$.
$f$ is continuous on $[a,x_1]$.
Intermediate Value Theorem:
There exists a point $p \in [a,x_1]$ with $f(p) =( f(a) + f(x_1))/2$.
Note:  $f(x_1) \lt f(p) \lt f(a)$.
We have $p \ne a$ ; $p\ne x_1$,
I.e. $a \lt p \lt x_1$.
Thus $f(p) \gt f(x_1)$,  where $p \lt x_1$, 
a contradiction to f being an increasing function.
Conclusion:  $f(a)$ is the minimum, hence $f(a) \le f(x)$ , for $x \gt a$.
