Let $E$ be a complex Hilbert space, $\mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$.
Consider the following two norms on $\mathcal{L}(E)\times \mathcal{L}(E)$ $$N_1:=N_1(A,B)=\left\|A^*A+B^*B\right\|^{1/2}$$ and $$N_2:=N_2(A,B)=\sup_{\substack{z_1,z_2\in \mathbb{C}^2,\\ |z_1|^2+|z_2|^2<1}}\left\|z_1A+z_2B\right\|.$$ It is not difficult to prove that $$N_2\leq N_1.$$
My goal is to find a constant $\alpha$ such that $$N_2\geq \alpha N_1.$$
It is easy to see that $F := \mathcal{L}(E) \times \mathcal{L}(E)$ is complete for $N_2$. Indeed, if $(A_n,B_n)$ is Cauchy for $N_2$ then by taking $z_1 = 1, z_2 = 0$ we see that $A_n$ is Cauchy in $\mathcal{L}(E)$ and hence converges to some $A \in \mathcal{L}(E)$. Similarly there is a $B$ such that $B_n \to B \in \mathcal{L}(E)$. It is then straightforward to see that $(A_n, B_n) \to (A,B)$ for $N_2$.
Now we check that $F$ is complete for $N_1$. To do this, I use the inequality you say you've proven. This implies that if $(A_n,B_n)$ is Cauchy for $N_1$ then it is Cauchy for $N_2$ and hence by the same argument as before, $A_n, B_n$ are both Cauchy in $\mathcal{L}(E)$. Hence, we are again in the situation where $A_n \to A$ and $B_n \to B$ in $\mathcal{L}(E)$. Then \begin{align} N_1(A_n - A, B_n - B)^2 &\leq \|(A_n - A)^* (A_n - A)\| + \|(B_n - B)^* (B_n - B)\| \\& = \|A_n - A\|^2 + \|B_n - B\|^2 \to 0 \end{align} as $n \to \infty$.
Hence $F$ is a Banach space under either of $N_1$ and $N_2$. The inequality you have then shows that $\operatorname{Id}: (F, N_1) \to (F,N_2)$ is a continuous linear bijection and hence by the open mapping theorem has continuous inverse. This implies the existence of the desired constant.