Let $\Omega\in\mathbb{R}^2$ be a bounded domain, $\forall a,b,c \in \mathbb{R}$, I hope to prove that there exists a positive constant $\alpha$ such that:
$$ \int_{\Omega} (ax_2+b)^2 + ( -ax_1+c)^2 \geqslant \alpha a^2 $$ where $\alpha$ is independent in $a,b,c$. I really hope for a directly proof instead of the proof by contradiction.
On
Hint: Expanding the squares on the left you get $a^{2}t+as+r$ with $t,r>0$ and $s \in \mathbb R$. What is left is to show that $(a^{2}t+as+r)/a^{2}$ has a positive minimum value. It is easy to compute the exact minimum of this expression and see that the minimum is positive if $s^{2} <4rt$. This inequality is an easy consequence of Cauchy-Schwarz inequality and I leave the details to you. [$bx_2+cx_1 \leq \sqrt {b^{2}+c^{2}} \sqrt {x_1^{2}+x_2^{2}}$]
Dividing by $a^2$, the integral is equal to \begin{equation} \int_\Omega \|X - M\|^2 d\mu(X) \end{equation} where $M = (c/a, -b/a)$ and $X = (x_1, x_2)$. This integral is minimal when $M$ is the center of mass of the domain \begin{equation} G = \frac{1}{\mu(\Omega)}\int_\Omega X d\mu(X) \end{equation} because $\|X-M\|^2 = \|X-G\|^2 + 2(X-G)\cdot(G-M) + \|G-M\|^2$ and $\int_\Omega (X - G)d\mu(X) = 0$, hence \begin{equation} \int_\Omega \|X - M\|^2 d\mu(X) = \int_\Omega \|X - G\|^2 d\mu(X) + \mu(\Omega)\|G - M\|^2 \end{equation} It follows that \begin{equation} \alpha = \int_\Omega \|X - G\|^2 d\mu(X) \end{equation}