How to prove $\left\|\ln\left(e^{iH_1}e^{iH_2}\right)\right\|\leq\left\|H_1\right\|+\left\|H_2\right\|$? Let $H_1$ and $H_2$ denote arbitrary Hermitian operators (finite dimensional) and let $\left\|\ldots\right\|$ denote the usual operator norm.  I conjecture that
$$
\left\|\ln\left(e^{iH_1}e^{iH_2}\right)\right\|\leq\left\|H_1\right\|+\left\|H_2\right\|\ ,
$$
but have no idea how to prove this. If $H_1$ and $H_2$ commute, it reduces to the triangle inequality. Cauchy-Schwarz and Golden-Thompson come to mind, but do not seem to help. 
 A: Given a unitary operator $U$ on $\mathbb{C}^n$, there exists an orthonormal basis $\{v_1,\dots,v_n\}$ and $\theta_1,\dots,\theta_n\in(-\pi,\pi]$, such that $Uv_k=e^{i\theta_k}v_k$, $k=1,\dots,n$. Let me define $\ln U$ be the unique operator which satisfies that $\ln U v_k=i\theta_kv_k$, $k=1,\dots,n$. With this definition, the conclusion is true. Note that $-i\ln U$ is Hermitian,  $\|\ln U\|=\max_{1\le k\le n}|\theta_k|\le \pi$, and $\|\ln U\|\le\|H\|$ for every  Hermitian operator $H$ with $U=e^{iH}$.
Let $\{e_1,\dots,e_n\}$ be a natural basis of $\mathbb{C}^n$ and let $\langle,\rangle$ be the standard inner product on $\mathbb{C}^n$, i.e. $\{e_1,\dots,e_n\}$ is an orthonormal basis of $\mathbb{C}^n$ w.r.t. $\langle,\rangle$. Since $\{e_1,\dots,e_n,ie_1,\dots,ie_n\}$ are $\mathbb{R}$-linearly independent, it gives a natural $\mathbb{R}$-linear isomorphism
$$f:\mathbb{C}^n\to\mathbb{R}^{2n},\quad z\mapsto (\mathrm{Re}z,\mathrm{Im}z)$$ Moreover, $\langle,\rangle$ induces an inner product $\langle,\rangle_{\mathbb{R}}$ on $\mathbb{R}^{2n}$ in the following way: given $z,w\in\mathbb{C}^n$, $\langle f(z),f(w)\rangle_{\mathbb{R}}:=\mathrm{Re}\langle z,w\rangle$. Then given two unit vectors $z,w\in\mathbb{C}^n$, we can define the angle between them as
$$\arccos\mathrm{Re}\langle z,w\rangle:=\angle(z,w)\in[0,\pi].$$
It is easy to show that for $\theta_k=\angle(z_k,w)$, $k=1,2$, $\angle(z_1,z_2)\le \theta_1+\theta_2$. To see this, firstly, since $\angle(z_1,z_2)\le\pi$, we may assume that $\theta_1+\theta_2\le\pi$. Secondly, up to a rotation, we may assume $w=e_1$. Then for $k=1,2$, $z_k=\cos\theta_ke_1+\sin\theta_kv_k$, where $\mathrm{Re}\langle e_1,v_k\rangle=0$ and $\|v_k\|=\sin\theta_k$. Therefore, Cauchy's inequality gives 
$$\cos\angle(z_1,z_2)\ge \cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2=\cos(\theta_1+\theta_2).$$
Therefore, $\angle(z_1,z_2)\le\theta_1+\theta_2$.
We claim that for any unitary operator $U$, 
$$\|\ln U\|=\sup_{\|z\|=1}\angle(z,Uz).$$ 
To see this, let $\theta_1,\dots,\theta_n\in(-\pi,\pi]$ be all the eigenvalues of $-i\ln U$. Then $\|\ln U\|=\max_{1\le k\le n}|\theta_k|$. Let $\{v_1,\dots,v_n\}$ be an orthonormal basis of $\mathbb{C}^n$, such that $Uv_k=e^{i\theta_k}v_k$,$k=1,\dots,n$. Now for any unit vector $z=\sum_{k=1}^na_kv_k$,
$$\mathrm{Re}\langle z,Uz\rangle=\mathrm{Re}\langle \sum_{k=1}^n a_kv_k,\sum_{k=1}^n a_ke^{i\theta_k}v_k\rangle=\sum_{k=1}^n|a_k|^2\cos\theta_k\ge\cos\|\ln U\|.$$
The equality can be realized by $z=v_k$ with $\theta_k=\|\ln U\|$, which completes the proof of the claim .
Now given Herimtian operators $H_1$ and $H_2$, denote $U_k=e^{iH_k}$, $k=1,2$ and $U=U_1U_2$. Since $U_1$ and $U_2$ are unitary, $U$ is unitary. Then 
$$\angle(z,Uz)=\angle(z,U_1U_2z)\le\angle(U_2z,U_1U_2z)+\angle(z,U_2z)\le \|\ln U_1\|+\|\ln U_2\|\le \|H_1\|+\|H_2\|.$$
Taking supremum on both sides, the conclusion follows.
