Finding $c\in\mathbb{R}_{>0}$ such that polynomial $p(x) \geq c (x-a)^n$. Let $p(x)$ be a polynomial of degree $n>1$, such that $p(x)>0$ for every $x>a$. Does there exist a $c\in\mathbb{R}_{>0}$ such that $p(x) \geq c(x-a)^n$ for all $x>a$?
In my first attempt I tried to use the binomial theorem. 
The right-hand side can be written as
$$c(x-a)^n = c\sum_{k=0}^n {n\choose k} x^k (-a)^{n-k}.$$
However, it became quite complicated with this approach, and I couldn't find a $c>0$.
In my second attempt I use division with rest, i.e. $p(x) = c(x-a)^n + r(x)$ for polynomial $r(x)$ with $\deg(r)<n$.
Let $p(x) = \sum_{k=0}^n a_k x^k$, we see that $a_n x^n$ should be equal to $cx^n$, because these are the only two expressions with degree $n$, so $c = a_n$.
Furthermore, $p(x)>0$ for $x>a$, so the leading coefficient $a_n > 0$, we now have a $c>0$ for which the equality holds. The only thing left to proof is that $r(x) \geq 0$.
First of all, is my second attempt correct? And how can I proof $r(x)\geq 0$ for $x>a$. Moreover, any other suggestions for $c>0$ are also welcome?
 A: Let first deal with the trivial case where $p(x)=k\,(x-a)^n$ for some $k\in\mathbb{R}$.  Then, $k>0$ and $c:=k$ clearly works (or set $c:=\dfrac{k}{2}$ if you want $p(x)>c\,(x-a)^n$ for all $x>a$).  From now on, we assume that $a$ is not a root of $p(x)$, or $a$ is a root of order less than $n$ of $p(x)$.
If $p(x)>0$ for all $x>a$, then the function $f:(a,\infty)\to\mathbb{R}$ defined by
$$f(x):=\frac{p(x)}{(x-a)^n}$$
for all $x>a$ is a positive continuous function (i.e., $f(x)>0$ for every $x>a$).  
Now, since $p$ is of degree $n$, we see that
$$\lim_{x\to\infty}\,f(x)=k\,,$$
where $k$ is the leading coefficient of $p$.  Obviously, $k>0$.   Thus, there exists $v>a$ such that $f(x)>\dfrac{k}{2}$ for every $x>v$.  
Furthermore, as $$\lim_{x\to a^+}\,f(x)=\infty\,,$$
there exists a real number $u$, $a<u<v$, such that $f(x)>k$ for every $x\in(a,u)$.  We can now look at $f$ on the interval $[u,v]$.
Note that $f|_{[u,v]}:[u,v]\to\mathbb{R}$ has a minimum $m>0$ (this is because $[u,v]$ is a compact interval).  Therefore, we clearly see that
$$f(x)>\min\left\{\frac{m}{2},\frac{k}{2}\right\}=:c$$
for every $x>a$.  This shows that
$$p(x)>c\,(x-a)^n$$
for all $x>a$.
You can prove a stronger result, using a similar proof to the one above.  Suppose that $p(x)$ and $q(x)$ are polynomials in $\mathbb{R}[x]$ such that $$p(x)>0\text{ for all }x> a\,.$$ If $p$ has degree greater than or equal to $q$, and that the order of $a$ as a root of $p$ is less than or equal to the order of $a$ as a root of $q$ (a root of order $0$ is simply a non-root), then there exists a constant $c>0$ such that $$p(x)\geq c\,q(x)$$ for all $x\geq a$.  You can choose $c$ so that the inequality is strict for $x>a$.
