= Solution
For a normalized trial state in the quadratic-form domain of a <self-adjoint> <Hamiltonian operator> bounded below, its <expectation> of $H$ is at least the ground <energy> $E_0$. Minimizing this <Rayleigh quotient> over a tractable trial family therefore supplies an upper bound.
For the two states, set $a_\pm=(a_1\pm a_2)/\sqrt2$. Their quotient is
$$
\frac{(E+\epsilon)a_+^2+(E-\epsilon)a_-^2}
{(1+s)a_+^2+(1-s)a_-^2}.
$$
The strict assumption $\epsilon<sE$ excludes $|s|=1$, since equality in the overlap Cauchy-Schwarz bound would then force the two states to be proportional and $\epsilon=sE$. Thus $|s|<1$. The two generalized <eigenvalues> are $(E\pm\epsilon)/(1\pm s)$, and their difference has the sign of $\epsilon-sE$. The symmetric trial is the lower one, so
$$
\boxed{E_0\leq\frac{E+\epsilon}{1+s}.}
$$
For the single attractive delta well, the supplied <wavefunction> has <norm> one because $\int\lambda e^{-2\lambda|x|}\,dx=1$. Its <kinetic energy> in quadratic-form notation is $\hbar^2\int|\psi_1'|^2/(2m)=\hbar^2\lambda^2/(2m)$, and its <delta potential> <energy> is $-K|\psi_1(0)|^2=-K\lambda$. With $\lambda=mK/\hbar^2$ this gives
$$
\boxed{E_B=-K\lambda/2.}
$$
Its derivative jump is $\psi_1'(0+)-\psi_1'(0-)=-2\lambda\psi_1(0)$, exactly the delta <boundary condition>, so it is an eigenstate. A decaying <bound state> on the two half-lines must have a common decay rate fixed by this same jump; there is only this one negative-energy state, confirming it is the <ground state>.
For $R>0$, integrate the overlap on the three regions $x<0$, $0<x<R$, $x>R$. Put $a=\lambda R$. The result is
$$
\boxed{s=(1+a)e^{-a},\qquad
E=E_B-K\lambda e^{-2a},\qquad
\epsilon=E_Bs-K\lambda e^{-a}.}
$$
The second equality is the single-well <energy> plus the extra delta <expectation>; the third follows by applying either single-well <eigenvalue> equation in the cross <matrix> element. Moreover $\epsilon-sE=K\lambda e^{-a}[(1+a)e^{-2a}-1]<0$, so the preceding bound applies. Substitution gives
$$
\boxed{E_0(R)\leq E_B\left[1+\frac{2e^{-\lambda R}(1+e^{-\lambda R})}{1+(1+\lambda R)e^{-\lambda R}}\right].}
$$
For widely separated wells, each localized orbital is close to an isolated-well <ground state> and the low-energy subspace is accurately their span; the symmetric combination captures the leading tunneling <energy> lowering. In the limit both exact and trial energies approach $E_B$. This illustrates electronic molecular binding: sharing an orbital between attractive centers can lower its <energy> below the separated-center value. A full equilibrium bond length would also require the repulsion of the positive centers.
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