Elliptic Operator Principal Part Inequality (Hints) I am looking to show the following estimate, for an elliptic PDO $P$ of order $m$ there is $K>0$, $R>0$ such that 
$$\bigg|\sum_{|\alpha| \leq m}a_{\alpha}\xi^\alpha \bigg| \geq K|\xi|^m$$
for all $\xi$ with $|\xi| > R$. I have shown the result for the principal part $P_m(\xi)$ and was looking to use that, but I cannot figure out how to do it for the full expression. My main idea was to take 
$$f(\xi) := \bigg|\sum_{|\alpha| \leq m}a_{\alpha}\xi^\alpha \bigg| - K|\xi|^m$$
and show that $f$ diverges to positive infinity as $|\xi|$ goes to $\infty$. This would mean that $f$ is bounded below, and also that it would eventually be positive which would give the result. Any hints on how to do any of this? 
 A: Here is an expanded version of the suggestion proposed in the comments by Hans Engler.
Consider any polynomial $Q(\xi) = \sum_{|\alpha| \le m} a_\alpha \xi^\alpha$ and note that $|Q(\xi)| \le \sum_{|\alpha| \le m} |a_\alpha| |\xi|^{|\alpha|}$.  Thus if we set 
$$
b_k = \sum_{|\alpha|=k} |a_\alpha|,
$$
then we can bound
$$
|Q(\xi)| \le \sum_{k=0}^m b_k |\xi|^k.
$$
Now set $\beta = \sum_{k=0}^m b_k$.  If $|\xi| \ge 1$ then 
$$
|Q(\xi)| \le \sum_{k=0}^m b_k |\xi|^k \le \sum_{k=0}^m b_k |\xi|^m = \beta |\xi|^m.
$$
On the other hand, if $|\xi |\le 1$ then 
$$
|Q(\xi)| \le \sum_{k=0}^m b_k |\xi|^k \le \sum_{k=0}^m b_k \cdot 1 = \beta.
$$
Consequently, 
$$
|Q(\xi)| \le \beta(1+|\xi|^m) \text{ for all } \xi \in \mathbb{R}^n.
$$
With this bound in hand we can apply it to your problem.  Write the symbol of your operator as $L = P + Q$ for $P = \sum_{|\alpha|=m} a_\alpha \xi^\alpha$ and $Q = \sum_{|\alpha| \le m-1} a_\alpha \xi^\alpha$.  You have already established that there exists $K >0$ such that
$$
|P(\xi)| \ge K |\xi|^m \text{ for all } \xi.
$$
Thus the triangle inequality tells us that
$$
|L(\xi)| = |P(\xi) + Q(\xi)| \ge |P(\xi)| - |Q(\xi)| \ge K |\xi|^m - \beta(1+|\xi|^{m-1}).
$$
Now we just observe that for $|\xi| > R$ with $R$ sufficiently large we must have that 
$$
\frac{K}{2} |\xi|^m - \beta(1+|\xi|^{m-1}) \ge 0,
$$
and so $|\xi| >R$ implies that
$$
|L(\xi)| \ge \frac{K}{2} |\xi|^m.
$$
