An inequality related to matrix norms and the trace distance of two matrices Suppose that $U_1,U_2$ are unitary matrices. A paper here (page 4) I am reading gives the following set of inequalities (using Dirac notation here for vectors representing pure quantum states, aka. tensor products of vectors in $\mathbb{C}^2$ with length $1$). 
$$
\|U_1 - U_2\| \geq \|U_1 |\psi \rangle - U_2 |\psi \rangle\| \geq \frac {1}{2}\operatorname{Trace}\left |U_1 |\psi \rangle \langle \psi| U_1^{\dagger} - U_2 |\psi \rangle \langle \psi| U_2^{\dagger}\right| \\
= D (U_1 |\psi \rangle \langle \psi| U_1^{\dagger} - U_2 |\psi \rangle \langle \psi| U_2^{\dagger}),$$
where $D$ is the trace distance. 
The norm used is not stated. Here are some things I know related to these inequalities:
1) Unitary matrices preserve the length of vectors
2) If $|| A||$ is a matrix norm induced by a vector norm, then for $||x||=1$, $||A|| \geq ||Ax||$, which would explain the first inequality. I'm not sure what is happening in the second inequality. Insights appreciated.
 A: The chain of inequalities makes sense if we add in the following.

Claim: For any unit vectors $x,y$, we have 
  $$
\|x - y\| \geq \frac 12 \operatorname{Tr}|xx^* - yy^*|.
$$

Proof: Noting that $x$ and $y$ are unit vectors, we compute
$$
\|x - y\|^2 = (x-y)^*(x-y) =  x^*x - x^*y - y^*x + y^*y \\
= 1 - \langle x, y\rangle - \langle y,x \rangle + 1 = 2(1 - \operatorname{Re}\langle x,y \rangle)
$$
So, we have $\|x - y\| = \sqrt{2}\sqrt{1 - \operatorname{Re}\langle x,y \rangle}$.
On the other hand, $A = (xx^* - yy^*)$ is a rank-2 Hermitian matrix with trace zero, which means that it has non-zero eigenvalues $\pm \lambda$ for some $\lambda>0$.  We compute
$$
\begin{align*}
    2\lambda^2 &= \lambda^2 + (-\lambda)^2 = \operatorname{Tr}(A^2) = \operatorname{Tr}[(xx^* - yy^*)^2] 
\\ & = \operatorname{Tr}[xx^*xx^* - xx^*yy^* - yy^*xx^* + yy^*yy^*]
    \\ & = 
    \operatorname{Tr}[x^*xx^*x - x^*yy^*x - y^*xx^*y + y^*yy^*y] 
\\ &= 
    \langle x,x\rangle^2 - 2|\langle x,y\rangle|^2 + \langle y,y \rangle^2
    \\ & = 
    2 - 2|\langle x,y\rangle|^2
\end{align*}
$$
We thereby conclude that $\lambda = \sqrt{1 - |\langle x,y \rangle|}$.  Thus, we compute
$$
\frac 12 \operatorname{Tr}|xx^* - yy^*| = \frac 12 (2 \lambda) = \sqrt{1 - |\langle x,y \rangle|}.
$$
Thus, it suffices to show that 
$$
\sqrt{2}\sqrt{1 - \operatorname{Re}\langle x,y \rangle} \geq 
\sqrt{1 - |\langle x,y \rangle|}.
$$
Indeed, we have
$$
\sqrt{2}\sqrt{1 - \operatorname{Re}\langle x,y \rangle} \geq 
\sqrt{1 - |\langle x,y \rangle|} \iff\\
2(1 - \operatorname{Re}\langle x,y \rangle) \geq 1 - |\langle x, y \rangle| \iff \\
1 + |\langle x, y \rangle| \geq 2 \operatorname{Re}\langle x,y \rangle \iff\\
\frac{1 + |\langle x, y \rangle|}{2} \geq \operatorname{Re}\langle x,y \rangle.
$$
The last inequality can be shown to hold as follows: by Cauchy Schwarz, $|\langle x, y \rangle| < 1$.  So, 
$$
\frac{1 + |\langle x, y \rangle|}{2} \geq |\langle x,y \rangle| = \sqrt{(\operatorname{Re}\langle x,y \rangle)^2 + (\operatorname{Im} \langle x,y\rangle)^2} \\
\qquad \qquad \quad\geq
\sqrt{(\operatorname{Re}\langle x,y \rangle)^2} = |\operatorname{Re}\langle x,y \rangle| \geq \operatorname{Re}\langle x,y \rangle.
$$
