Define $\mathbb{W} = H_{0}^{1}(-1,1) \times H_{0}^{1}(-1,1)$, where $u \in H_{0}^{1}(-1,1)$ if $u,u^{\prime} \in L^{2}(-1,1)$ and $u(-1) = u(1) = 0$. Consider the sesquilinear form $a: \mathbb{W} \times \mathbb{W} \to \mathbb{C}$ given by $$ a((u,v),(w,z)) = \int_{-1}^{1}u^{\prime}\overline{w}^{\prime}dx + \int_{-1}^{1}v^{\prime}\overline{z}^{\prime}dx + \int_{-1}^{1}u\overline{z}dx + \int_{-1}^{1}v\overline{w}dx $$ for $(u,v),(w,z) \in \mathbb{W}$. I want to show that $a$ is coercive, that is, there is $C > 0$ such that
$$ \text{Re}~a((u,v),(u,v)) \geq C \|(u,v)\|_{W}^{2}, \ \ \forall (u,v) \in \mathbb{W}. $$
Where $\|(u,v)\|_{W} = \|u\|_{H_{0}^{1}} + \|v\|_{H_{0}^{1}}$, such that $\|u\|_{H_{0}^{1}} = \|u^{\prime}\|_{L^{2}}^{2} + \|u\|_{L^{2}}^{2} $
Let $(u,v) \in \mathbb{W}$ and
$$
a((u,v),(u,v)) = \int_{-1}^{1}|u^{\prime}|^{2}dx + \int_{-1}^{1}|v^{\prime}|^{2}dx + \int_{-1}^{1}u\overline{v}dx + \int_{-1}^{1}v\overline{u}dx
$$
But
$$
\int_{-1}^{1}u\overline{v}dx + \int_{-1}^{1}v\overline{u}dx = 2\text{Re}\int_{-1}^{1}u\overline{v}dx
$$
Then, by Poincaré’s inequality
$$
a((u,v),(u,v)) = \|u^{\prime}\|_{L^{2}}^2 + \|v^{\prime}\|_{L^{2}}^2 + 2\text{Re}\int_{-1}^{1}u\overline{v}dx \geq 2^{-3/2}\|u\|_{L^{2}(-1,1)}^2 + 2^{-3/2}\|v\|_{L^{2}(-1,1)}^2 + 2\text{Re}\int_{-1}^{1}u\overline{v}dx
$$
I know that
$$
0 \leq (\|u\|_{L^{2}} - \|v\|_{L^{2}})^{2} =
(\|u\|_{L^{2}}^2 + \|v\|_{L^{2}}^2) - 2\|u\|_{L^{2}} \|v\|_{L^{2}}
$$
Then
$$
2\text{Re}\int_{-1}^{1}u\overline{v}dx \leq 2\bigg|\int_{-1}^{1}u\overline{v}dx \bigg| \leq 2\|u\|_{L^{2}} \|v\|_{L^{2}} \leq \|u\|_{L^{2}}^2 + \|v\|_{L^{2}}^2
$$
But I need a contrary inequality and not the one above. Help me, please!!
