I am reading a proof of some properties of quasi-central approximate unit. At some point the author claims that is using the following inequality, valid for any families $(a_j)_{j\in J}$, $(b_j)_{j\in J}$, indexed by a given finite set $J$, in a given $C^\ast$-algebra $A$: $$\left\|\sum\limits_{j\in J} a_jb_j\right\|\leq \left\|\sum\limits_{j\in J}a_ja_j^\ast\right\|^{1/2}\left\|\sum\limits_{j\in J}b_j^\ast b_j\right\|^{1/2}.$$
I do not see how exactly it is used in this specific situation, so let me briefly summarize what was done, and then I will formulate my question. First, we have an ideal $\mathcal{I}\subset A$, and a non-decreasing net $(\sigma_i)$ in the unit ball of $\mathcal{I}$. Moreover, $\sigma_i\geq 0$. It satisfied some futher conditions but they are irrelevant for what I need now. Take some $a,b\in A$. It is claimed that the above inequality immediately implies the following one: $$\|\sigma_i^{1/2}a\sigma_i^{1/2}+(1-\sigma_i)^{1/2}b(1-\sigma_i)^{1/2}\|\leq \max\{\|a\|,\|b\|\}.$$ Why it is the case? Maybe I do not see something obvious, but I couldn't find a way how to use the previous inequality to deduce this one.
Suppose that $\|\sigma\|\leq1$. Take $$a_1=\sigma^{1/2},\ \ a_2=(1-\sigma)^{1/2}, \ \ b_1=a\sigma^{1/2},\ \ b_2=b(1-\sigma)^{1/2}.$$
Then, using the inequality in the first line, \begin{align} \|\sigma^{1/2}a\sigma^{1/2}+(1-\sigma)^{1/2}b(1-\sigma)^{1/2}\| &\leq\|\sigma+1-\sigma\|^{1/2}\,\|\sigma^{1/2}a^*a\sigma^{1/2}+(1-\sigma)^{1/2}b^*b(1-\sigma)^{1/2}\|^{1/2}\\[0.3cm] &=\|\sigma^{1/2}a^*a\sigma^{1/2}+(1-\sigma)^{1/2}b^*b(1-\sigma)^{1/2}\|^{1/2}\\[0.3cm] &\leq\big\|\sigma^{1/2}\,\|a^*a\|\,\sigma^{1/2}+(1-\sigma)^{1/2}\,\|b^*b\|\,(1-\sigma)^{1/2}\big\|^{1/2}\\[0.3cm] &=\big\|\sigma^{1/2}\,\|a\|^2\,\sigma^{1/2}+(1-\sigma)^{1/2}\,\|b\|^2\,(1-\sigma)^{1/2}\big\|^{1/2}\\[0.3cm] &\leq\max\{\|a\|\,\|b\|\}\,\|\sigma+1-\sigma\|^{1/2}\\[0,3cm] &=\max\{\|a\|,\|b\|\}. \end{align} All that's used in the remaining inequalities is that if $0\leq x\leq y$, then $\sigma x\sigma^*\leq\sigma y\sigma^*$, and $\|x\|\leq\|y\|$.