= Solution
For a normalized trial function in the quadratic-form domain of the <Schrödinger operator>, integration by parts gives $\langle E\rangle=\langle\psi,H\psi\rangle$. Expand in the complete orthonormal <energy eigenstates>, $\psi=\sum_nc_n\psi_n$ with $\sum_nc_n^2=1$. Then
$$
\langle E\rangle=\sum_nE_nc_n^2\geq E_0\sum_nc_n^2=E_0.
$$
Minimizing over a trial family therefore gives the best upper bound within that family, the <quantum variational principle>.
For $V=|x|$, choose $\psi_\alpha=(\alpha/\pi)^{1/4}e^{-\alpha x^2/2}$ with $\alpha>0$. The Gaussian integrals give
$$
\int(\psi_\alpha')^2dx=\frac\alpha2,\qquad
\int|x|\psi_\alpha^2dx=\frac1{\sqrt{\pi\alpha}}.
$$
Thus $E(\alpha)=\alpha/2+(\pi\alpha)^{-1/2}$, whose derivative vanishes at $\alpha=\pi^{-1/3}$. It tends to infinity at both parameter endpoints and has positive second derivative, so this is its minimum. \b[$\boxed{E_0\leq\tfrac32\pi^{-1/3}\simeq1.0242}$] is the resulting estimate.
The potential is even; the unique nodeless ground state is even and the first excited state is odd, by the one-dimensional <Sturm oscillation theorem>. To estimate $E_1$, minimize the same <Rayleigh quotient> over odd trial functions, for example normalized $x e^{-\alpha x^2/2}$, or the compactly supported odd family $C_a x(1-|x|/a)$ for $|x|\leq a$ and zero otherwise. Oddness gives exact orthogonality to the true ground state, not merely to an approximate trial ground state. The <quantum variational principle> restricted to the odd sector supplies an upper bound for $E_1$. No further integration is needed.
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