Let $A$ and $B$ be $C$*-algebras and $\phi:A\to B$ a completely positive contractive map. I want to show that, for any $a,b\in A$; $$\Vert \phi(ab)-\phi(a)\phi(b)\Vert\leq\Vert\phi(aa^*)-\phi(a)\phi(a^*)\Vert^\frac{1}{2}\Vert b\Vert.$$
After reading some solutions for inequalities related to completely positive maps, it seems Stinespring's theorem and the $2$-positive trick are good starting points but I don't know how to proceed. A related post containing the mentioned ideas is here. Indeed, $\phi(ab)-\phi(a)\phi(b)$ is very similar to the determinant of a martix, for example, $ \begin{bmatrix} \phi(ab) & \phi(a) \\ \phi(b) & I \\ \end{bmatrix} $.
Any help would be highly appreciated.
Let $$X=\begin{bmatrix} a^*&b\\0&0\end{bmatrix}.$$ Because $\phi$ is ccp (though $2$-positive and contractive is enough), $$ \phi^{(2)}(X)^*\phi^{(2)}(X)\leq\phi^{(2)}(X^*X). $$ This gives us $$ \begin{bmatrix} \phi(aa^*)-\phi(a)\phi(a)^*&\phi(ab)-\phi(a)\phi(b)\\ [\phi(ab)-\phi(a)\phi(b)]^*&\phi(b^*b)-\phi(b)^*\phi(b) \end{bmatrix}\geq0. $$ It is not hard to show that a matrix $\begin{bmatrix} r&s\\ s^* &t\end{bmatrix} $ is positive if and only if for every $\varepsilon>0$ there exists a contraction $x_\varepsilon$ such that $s=(r+\varepsilon)^{1/2}x_\varepsilon (t+\varepsilon)^{1/2}$. Thus $$ \|\phi(ab)-\phi(a)\phi(b)\|\leq\|\varepsilon+\phi(aa^*)-\phi(a)\phi(a)^*\|^{1/2}\,\|\varepsilon+\phi(b^*b)-\phi(b)^*\phi(b)\|^{1/2}. $$ Since $\phi(b^*b)-\phi(b)^*\phi(b)\geq0$ and $\phi(b^*)\phi(b)\geq0$, we have $$ \|\phi(b^*b)-\phi(b)^*\phi(b)\|\leq\|\phi(b^*b)\|\leq\|b\|^2. $$ This, together with letting $\varepsilon\to0$, gives us $$ \|\phi(ab)-\phi(a)\phi(b)\|\leq\|\phi(aa^*)-\phi(a)\phi(a)^*\|^{1/2}\,\|b\| $$