We are given an elliptic operator $P=\sum_{|\alpha|\leq m}a_\alpha\partial^\alpha$ that is elliptic in $\Omega$. $a_\alpha$ are constants. I am supposed to show that $$\|u\|_s\leq C_s(\|u\|_0+\|Pu\|_{s-m}).$$ I believe the definition of elliptic we're supposed to use is that there exists $\gamma>0$ such that $$\left|\sum_{|\alpha|=m}a_\alpha(x)\xi^\alpha\right|\geq c|\xi|^m$$ for all $\xi\in\mathbb R^n$ and $x\in\Omega$.
I believe this is supposed to be an easy problem but I'm stuck.
My attempt so far is as follows:
I think we can write using the reverse triangle inequality \begin{align*} |\widehat{Pu}(\xi)|&=|\sum_{|\alpha|\leq m}a_\alpha\widehat{\partial^\alpha u}(\xi)|\\ &=|\sum_{|\alpha|\leq m}a_\alpha(i\xi)^\alpha\hat u(\xi)|\\ &\geq C\left(|\sum_{|\alpha|=m}a_\alpha(i\xi)^\alpha\hat u(\xi)|-|\sum_{|\alpha|<m}a_\alpha(i\xi)^\alpha\hat u(\xi)|\right)\\ &\geq C\left(\gamma|\xi|^m|\hat u(\xi)|-\sum_{|\alpha|<m}|a_\alpha\xi^m\hat u(\xi)|\right) \end{align*} which might give something like \begin{align*} \|Pu\|_{s-m}&=\int_\Omega(1+|\xi|^2)^{s-m}|\widehat{Pu}(\xi)|^2d\xi\\ &\geq C\int_\Omega(1+|\xi|^2)^{s-m}\gamma|\xi|^m|\hat u(\xi)|^2d\xi-C\sum_{|\alpha|<m}\int_\Omega(1+|\xi|^2)^{s-m}|\xi|^m|\hat u(\xi)|^2d\xi. \end{align*} The first term is then bounded by $C\|u\|_s$ but I don't know how to bound the second term by $C\|u\|_0$.
Using the reverse triangle inequality the way you did, too much stuff goes into the subtracted term. You need a more careful estimate, namely $$ \left|\sum_{|\alpha|\le m} a_\alpha(x)\xi^\alpha\right|\geq \frac{\gamma}{2}|\xi|^m - B \tag{1} $$ where $\gamma$ is from the definition of ellipticity and $B$ is a constant. When $(1)$ is applied to estimate $\|P_u\|_{s-m}$ via the Fourier transform, the result follows.
To prove $(1)$, observe that $$\xi \mapsto \left|\sum_{|\alpha|\le m} a_\alpha(x)\xi^\alpha\right| - \frac{\gamma}{2}|\xi|^m$$ is a continuous function of $\xi$ that tends to $\infty$ as $|\xi|\to\infty$; therefore, it is bounded below.