= Solution
The <Rayleigh quotient> is $R[\psi]=\langle\psi,H\psi\rangle/\langle\psi,\psi\rangle$, interpreted through the energy <quadratic form> for trial states in its form domain. Expanding in normalized energy eigenstates gives $R[\psi]=\sum E_n|c_n|^2/\sum|c_n|^2\geq E_0$. If the ground state is unique, equality requires all coefficients outside its one-dimensional <eigenspace> to vanish. Thus equality means the same ground-state ray, allowing normalization and an overall phase.
Use the normalized trial function $\sqrt{2/a}\cos(\pi x/a)$ on $|x|<a/2$, zero outside. Its <kinetic energy> is $\pi^2\hbar^2/(2ma^2)$ and its mean square position is $c_0a^2$, where $c_0=1/12-1/(2\pi^2)$. The energy is
$$
R_0(a)=\frac{\pi^2\hbar^2}{2ma^2}+\frac{m\omega^2c_0a^2}{2}.
$$
Minimization gives $a^4=\pi^2\hbar^2/(m^2\omega^2c_0)$ and
$$
\boxed{E_0\leq\hbar\omega\sqrt{\frac{\pi^2}{12}-\frac12}\approx0.56786\hbar\omega}.
$$
For the first excited well state use $\sqrt{2/a}\sin(2\pi x/a)$, zero outside. Now the <kinetic energy> is $2\pi^2\hbar^2/(ma^2)$ and the position factor is $c_1=1/12-1/(8\pi^2)$. Minimizing gives $a^4=4\pi^2\hbar^2/(m^2\omega^2c_1)$ and
$$
\boxed{E_1\leq\hbar\omega\sqrt{\frac{\pi^2}{3}-\frac12}\approx1.67029\hbar\omega}.
$$
\b[This particular excited-state estimate is an upper bound.] The trial function is odd and the oscillator ground state is even, so it is exactly orthogonal to the ground state. Its spectral expansion therefore contains only energies at least $E_1$, giving the same variational bound on that orthogonal subspace. An arbitrary trial state without this <orthogonality> would bound only $E_0$, not $E_1$. The compactly supported trial functions belong to the quadratic-form domain despite <derivative> jumps at the well edges; their kinetic energies are correctly computed as $\hbar^2\int|\psi'|^2/(2m)$.
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