$\mathcal{H}$ : real Hilbert space with inner product $(\,,\,)$ and norm $||\,||:=(\,,\,)^{1/2}$
Let $D$ be a linear subspace of $\,\mathcal{H}$ and $\mathcal{E}$ : $D\times D\to \mathbb{R}$ a bilinear map. For $\alpha\geq0$ we set
$\mathcal{E}_{\alpha}(u,v):=\mathcal{E}(u,v)+\alpha(u,v)\quad(\,u,v \in D \times D \,)$
Assume $\mathcal{E}(u,u)\geq 0$ for all $ u \in D $ and $\mathcal{E}(u,u)=0 \Leftrightarrow u=0 $
I want to prove that the following assertions are all equivalent:
$(a)$ $\exists K>0$ s.t. $|\mathcal{E}_{1}(u,v)|\leq K \mathcal{E}_{1}(u,u)^{1/2}\mathcal{E}_{1}(v,v)^{1/2}\quad(\,\forall u, \forall v \in D\,)$
$(b)$ $\forall \alpha>0,\,\exists K_{\alpha}>0$ s.t. $|\mathcal{E}_{\alpha}(u,v)|\leq K_{\alpha} \mathcal{E}_{\alpha}(u,u)^{1/2}\mathcal{E}_{\alpha}(v,v)^{1/2}\quad(\,\forall u,\forall v \in D\,)$
$(c)$ $\forall \alpha>0,\,\exists K_{\alpha}'>0$ s.t. $|\mathcal{E}(u,v)|\leq K_{\alpha}' \mathcal{E}_{\alpha}(u,u)^{1/2}\mathcal{E}_{\alpha}(v,v)^{1/2}\quad(\,\forall u,\forall v \in D\,)$
$(b) \Rightarrow (a)$ is clear. Suppose $(a) \Rightarrow (c)$ is proved, I think assertion $(c) \Rightarrow (b)$ is similarly proved.
Therefore, I tried to prove assertion $(a) \Rightarrow (c) $ as follows:
By definition $\mathcal{E}_{1}(u,v)=\mathcal{E}(u,v)+(u,v)$, assertion (a) can be transformed as follows:
$|\mathcal{E}(u,v)|\leq K \mathcal{E}_{1}(u,u)^{1/2}\mathcal{E}_{1}(v,v)^{1/2}+|(u,v)|$
Also by definition $\mathcal{E}_{\alpha}(u,v)=\mathcal{E}(u,v)+\alpha(u,v)$ and then
$|\mathcal{E}(u,v)|\leq K \mathcal{E}_{1}(u,u)^{1/2}\mathcal{E}_{1}(v,v)^{1/2}+ \left| \frac{\mathcal{E}_{\alpha}(u,v)-\mathcal{E}(u,v)}{\alpha} \right|$
$\left(1-\frac{1}{\alpha}\right) |\mathcal{E}(u,v)|\leq K \mathcal{E}_{1}(u,u)^{1/2}\mathcal{E}_{1}(v,v)^{1/2}+\frac{1}{\alpha}| \mathcal{E}_{\alpha}(u,v)| $
Since $\mathcal{E}_{\alpha}(u,v)$ is inner product, from Cauchy-Schwarz inequality
$\left(1-\frac{1}{\alpha}\right) |\mathcal{E}(u,v)|\leq K\mathcal{E}_{1}(u,u)^{1/2}\mathcal{E}_{1}(v,v)^{1/2}+\frac{1}{\alpha}\mathcal{E}_{\alpha}(u,u)^{1/2}\mathcal{E}_{\alpha}(v,v)^{1/2} $
If $\alpha>1$ then $\mathcal{E}_{1}(u,u)\leq \mathcal{E}_{\alpha}(u,u)$ and $\left(1-\frac{1}{\alpha}\right) |\mathcal{E}(u,v)|\leq \left(K+\frac{1}{\alpha}\right)\mathcal{E}_{\alpha}(u,u)^{1/2}\mathcal{E}_{\alpha}(v,v)^{1/2} $
I can't deal with the case of $\,0<\alpha<1$. Please teach me.
The form ${\cal E}_\alpha(u,v)$ is the sum of two forms: ${\cal E}(u,v)$ which is not symmetric, and $\alpha(u,v)$ which is symmetric. Now, let's exploit Cauchy-Schwarz on the symmetric part to obtain
$$|\alpha(u,v)|\leq \alpha (u,u)^{1/2}\,(v,v)^{1/2}\leq {\cal E}_\alpha^{1/2}(u,u)\,{\cal E}_\alpha^{1/2}(v,v). $$ You should be able to take it from here.