= Solution
Here is a finite-dimensional form of the <Kitaev geometrical lemma>. Let $A,B$ be <positive operators>, with nullspaces $S_A,S_B$ and all positive <eigenvalues> at least $\gamma>0$. First assume $S_A\cap S_B=\{0\}$. Let $P,Q$ be the <orthogonal projections> onto those nullspaces, and define their <smallest angle between two subspaces> by
$$
\cos\vartheta=
\sup_{\substack{u\in S_A,\ v\in S_B\\\|u\|=\|v\|=1}}
|\langle u,v\rangle|
=\|PQ\|,\qquad 0<\vartheta\leq\pi/2.
$$
If either nullspace is zero, set $\cos\vartheta=0$. Then
$$
\boxed{A+B\geq
\gamma(1-\cos\vartheta)I
=2\gamma\sin^2(\vartheta/2)I.}
$$
To prove it, spectral decomposition gives $A\geq\gamma(I-P)$ and $B\geq\gamma(I-Q)$. Write $C=P+Q$ and $c=\cos\vartheta$. For every vector $z$,
$$
\begin{aligned}
\langle z,C^2z\rangle
&=\|Pz\|^2+\|Qz\|^2+2\operatorname{Re}\langle Pz,Qz\rangle\\
&\leq\|Pz\|^2+\|Qz\|^2+2c\|Pz\|\|Qz\|\\
&\leq(1+c)(\|Pz\|^2+\|Qz\|^2)
=(1+c)\langle z,Cz\rangle.
\end{aligned}
$$
Thus $C^2\leq(1+c)C$. Since $C$ is <positive semidefinite>, each of its <eigenvalues> lies in $[0,1+c]$, so $2I-C\geq(1-c)I$. Consequently
$$
A+B\geq\gamma(2I-P-Q)\geq\gamma(1-c)I,
$$
as claimed.
If there is a common nullspace $S=S_A\cap S_B$, restrict to $S^\perp$ and define the angle between $S_A\cap S^\perp$ and $S_B\cap S^\perp$ there. The same proof bounds the smallest positive <eigenvalue> of $A+B$ by $2\gamma\sin^2(\vartheta/2)$, while $S$ is its zero-energy subspace. This version gives a <spectral gap> above a possibly degenerate <ground state> space.
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