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,
$$
\frac{1 + |\langle x, y \rangle|}{2} \geq |\langle x,y \rangle| = \sqrt{(\operatorname{Re}\langle x,y \rangle)^2 + (\operatorname{Im} \langle x,y\rangle)^2} \\
\qquad \qquad \quad\geq
\sqrt{(\operatorname{Re}\langle x,y \rangle)^2} = |\operatorname{Re}\langle x,y \rangle| \geq \operatorname{Re}\langle x,y \rangle.
$$
Best Answer
Assume that a mapping $H\mapsto U$ is of such a form that $U$ can be expressed as a power series in $H$ (I don't think this is a huge loss of generality to assume this, perhaps none.) Then $$ U=c_{0}I+c_{1}H+c_{2}H^{2} + c_{3}H^{3} + \cdots $$ Since $U$ unitary implies $e^{i\theta}U$ is unitary, then WLOG we may take $c_{0}$ to be real. Moreover, if $U$ is unitary, then $U^{-1}= U^{*}$, so (assuming $U^{-1}$ can be represented as a power series in $H$) $$ U^{-1}=c_{0}I+c_{1}^{*}H+c_{2}^{*}H^{2} + c_{3}^{*}H^{3}+ \cdots $$ The condition $UU^{-1} = I$ gives $$ I=UU^{-1}=c_{0}^{2}I+c_{0}(c_{1}+c_{1}^{*})H+\cdots $$ whereupon we immediately determine that $c_{0}=1$ and that $c_{1}+c_{1}^{*}=0$. Hence $c_{1}$ is purely imaginary, and we write $c_{1}=ib_{1}$. With this in hand, we expand the products of these power series again: $$ I=UU^{-1}=I+(c_{2}+c_{2}^{*}+b_{1}^{2})H^{2}+\cdots $$ so that $2\mathfrak{R}(c_{2})=-b_{1}^{2}$ but the imaginary part of $c_{2}$ is a free parameter. So we've already found 2 free parameters, each of which generate infinite families of unitary maps $H\mapsto U$. So there are a ton of these maps.