Let $A_1$ and $A_2$ two bounded linear operators on a complex Hilbert space $E$.
I want to prove that $$\frac{1}{2}N_1(A_1,A_2)\leq N_2(A_1,A_2) \leq N_1(A_1,A_2),$$ where $$N_1(A_1,A_2)=\displaystyle\sup\left\{\|A_1 x\|^2+\|A_2 x\|^2;\;x,y\in E,\;\|x\|=1\right\},$$ and $$N_2(A_1,A_2)=\displaystyle\sup\left\{|\langle A_1 x, y \rangle|^2+|\langle A_2 x, y \rangle|^2;\;x,y\in E,\;\|x\|=\|y\|=1\right\}.$$
I think that the inequality $N_2(A_1,A_2) \leq N_1(A_1,A_2)$ follows by applying the C.S. inequality. But I cannot prove the first inequality.
For $\|x\|= 1$ and $y_1=\displaystyle {A_1x\over \|A_1x\|},$ $\displaystyle y_2={A_2x\over \|A_2x\|},$ we have $$|\langle A_1x,y_1\rangle |^2=\|A_1x\|^2\qquad |\langle A_2x,y_2\rangle |^2=\|A_2x\|^2$$ (if $A_jx=0$ we set $y_j=x$). Thus we get $$N_2(A_1,A_2)\ge |\langle A_1x,y_1\rangle |^2+|\langle A_2x,y_1\rangle |^2\ge \|A_1x\|^2$$ $$N_2(A_1,A_2)\ge |\langle A_1x,y_2\rangle |^2+|\langle A_2x,y_2\rangle |^2\ge \|A_2x\|^2$$
Hence $$N_2(A_1,A_2)\ge {1\over 2}[\|A_1x\|^2+\|A_2x\|^2]$$ and $$N_2(A_1,A_2)\ge {1\over 2}\sup_{\|x\|=1}[\|A_1x\|^2+\|A_2x\|^2]= {1\over 2}N_1(A_1,A_2).$$
The inequalities are sharp. It can be seen already in two dimensions. Let $$A_1=\begin{pmatrix} 1& 0 \\ 0 & 0 \end{pmatrix},\qquad A_2=\begin{pmatrix} 0& 0 \\ 1 & 0 \end{pmatrix}$$ Then $$N_1(A_1,A_2)= 2,\qquad N_2(A_1,A_2)=1$$ The first equality is attained at $x=(1,0).$ Concerning the second one observe that for $x=(x_1,x_2)$ and $y=(y_1,y_2)$ we get $$\displaylines{|\langle A_1x,y\rangle |^2+|\langle A_2x,y\rangle |^2\\ =|x_1|^2|y_1|^2+|x_1|^2|y_2|^2 \le 1}$$ if $\|x\|=\|y\|=1.$