I was trying to prove Remark 3.7 in this article, but failed. Here some necessary definitions.
Let $\mathcal{A}$ be a unital $C^*$-algebra such that there exist $S_1,S_2\in\mathcal{A}$ with property $S_1^*S_1=S_2^*S_2=1_{\mathcal{A}}$, $S_1^*S_2=S_2^*S_1=0$ (i.e. isometries with pairwise orthogonal images)
Let $Y$ be contractive (i.e. for all $a\in\mathcal{A}$, $y\in Y$ we have $\Vert a\cdot y\Vert\leq\Vert a\Vert\Vert y\Vert$) normed (i.e. normed space) unital (i.e. for all $y\in Y$ we have $1_\mathcal{A}y=y$) left module over $\mathcal{A}$.
Such a normed module is called semi-Ruan if for all $y_1, y_2\in Y$ the following holds $ \Vert S_1\cdot y_1+S_2\cdot y_2\Vert^2\leq\Vert y_1\Vert^2+\Vert y_2\Vert^2 $
My question. Let $Y^*$ be dual module of semi-Ruan module $Y$ (i.e. dual space with the structure of right unital contractive normed module with outer multiplication defined by $(f\cdot a)(y)=f(a\cdot y)$ ), then the following inequality holds $ \Vert f_1\cdot S_1^*+f_2\cdot S_2^*\Vert^2\geq\Vert f_1\Vert^2+\Vert f_2\Vert^2 $ My attempt. Note that $\Vert y_1\Vert\leq 1$,$\Vert y_2\Vert\leq 1$ implies $\Vert S_1\cdot y_1+S_2\cdot y_2\Vert\leq\sqrt{2}$, so $ \Vert f_1\cdot S_1^*+f_2\cdot S_2^*\Vert^2=\sup\{|(f_1\cdot S_1^*+f_2\cdot S_2^*)(y)|^2:\Vert y\Vert\leq 1\}\geq $ $ \sup\left\{\left|(f_1\cdot S_1^*+f_2\cdot S_2^*)\left(S_1\cdot\frac{y_1}{\sqrt{2}}+S_2\cdot\frac{y_2}{\sqrt{2}}\right)\right|^2:\Vert y_1\Vert\leq 1,\Vert y_2\Vert\leq 1\right\}= $ $ 0.5 \sup\left\{|f_1(y_1)+f_2(y_2)|^2:\Vert y_1\Vert\leq 1,\Vert y_2\Vert\leq 1\right\}= $ $ 0.5 \sup\left\{|f_1(y_1)|^2+|f_2(y_2)|^2+2\operatorname{Re}(f_1(y_1)\overline{f_2(y_2)}):\Vert y_1\Vert\leq 1,\Vert y_2\Vert\leq 1\right\}=??? $