I'm studing the Brezzi Theorem (also called Babuška-Brezzi or Ladyzhenskaya-Babuška-Brezzi Theorem).
The Theorem 1 (on that link) ensures the continuous dependence of the solution, that is
$$\|(u,\lambda)\|_{X\times M}\leq C\{\|f\|_{X^*}+\|g\|_{M^*}\}.$$
With this expresion and recalling that
$$\|f\|_{X^*}=\sup_{v\in X}\dfrac{\langle f,v\rangle}{\|v\|_X}\quad\text{ and }\quad\|g\|_{M^*}=\sup_{\mu\in M}\dfrac{\langle g,\mu\rangle}{\|\mu\|_M}$$
we obtain that
$$\|(u,\lambda)\|_{X\times M}\leq C\left\{\sup_{v\in X}\dfrac{\langle f,v\rangle}{\|v\|_X}+\sup_{\mu\in M}\dfrac{\langle g,\mu\rangle}{\|\mu\|_M}\right\}$$
$$=C\left\{\sup_{v\in X}\dfrac{a(u,v)+b(v,\lambda)}{\|v\|_X}+\sup_{\mu\in M}\dfrac{b({\color{red}u},\mu)}{\|\mu\|_M}\right\}$$
My question is, it is possible to prove the following?
$$\boxed{\|(u,\lambda)\|_{X\times M}\leq C\left\{\sup_{(v,\mu)\in X\times M}\dfrac{a(u,v)+b(v,\lambda)+b({\color{red}u},\mu)}{\|(v,\mu)\|_{X\times M}}\right\}}$$
This is something like a global inf-sup condition.