An inequality for completely positive maps. Let $f\colon A\to B$ be a contractive completely positive, ${}^*$-preserving map between C*-algebras and take $a\in A$. How one can prove that
$$0\leqslant f(a)f(a^*)\leqslant f(aa^*)?$$
Some authors take it for granted without any explanation.
 A: Just for fun, the matrix trick mentioned in the answer by @Tom Cooney:
Assume $A$ is unital. If not, extend $f$ to unitization of $A$.
For any $a \in A$, 
$$
\left[
\begin{matrix}
1 & a^*\\
a & aa^*
\end{matrix}
\right]
$$ 
is positive in $M_2(A)$. The $2$-positivity of $f$ then tells you
$$
f(a) f(a^*) \leq f(aa^*).
$$
A: One of the most important results about completely positive maps is Stinespring's Dilation Theorem.
Suppose that $f:A \to B$ is a completely positive map, where $A$ and $B$ are $C^*$-algebras. Then we can find a Hilbert space $H$ such that $B \subseteq B(H)$.
Stinespring's Theorem then states that there exists a Hilbert space $K$,  $*$-homomorphism $\pi: A \to B(K)$, and a bounded operator $V: H \to K$ such that 
$$
f(a)= V \pi(a) V^*,
$$
with $\Vert f \Vert = \Vert V \Vert^2$.
The desired inequality now follows easily:
\begin{array}{ccc}
f(aa^*)& =& V \pi(aa^*) V^* \\
   &= &V \pi(a) \pi(a^*) V^* \\
   & \geq &V \pi(a) V^* V \pi(a^*) V^* \\
    &= &f(a) f(a^*).
\end{array}
Here we use that $ V^*V \leq \Vert V \Vert^2 1 \leq \Vert f \Vert \leq 1$ because $f$ is contractive.
In fact, one can prove this result only using the weaker hypothesis that $f$ is $2$-positive. This is known as Kadison's Inequality.
