= Solution
Use the real <weighted inner product> $\langle u,v\rangle_w=\int_a^bwuv\,dx$ and the energy bilinear form $B(u,v)=\int_a^b(pu'v'+quv)dx$. <Integration by parts>, the endpoint conditions, and the <Sturm-Liouville problem> give
$$
B(y_i,y_j)=\int_a^by_i[-(py_j')'+qy_j]dx=\lambda_j\langle y_i,y_j\rangle_w=\lambda_j\delta_{ij}.
$$
Thus <orthonormality> in the stated expansion gives
$$
F[y_{\rm trial}]=A^2\left(\lambda_0+\sum_{i\geq1}c_i^2\lambda_i\right),\qquad G[y_{\rm trial}]=A^2\left(1+\sum_{i\geq1}c_i^2\right),
$$
and hence the <Rayleigh quotient>
$$
\boxed{\lambda_{\rm trial}=\frac{\lambda_0+\sum_{i\geq1}c_i^2\lambda_i}{1+\sum_{i\geq1}c_i^2}.}
$$
The expansion is understood for nonzero admissible functions with finite norm and energy.
For a general endpoint-preserving variation $y+\varepsilon\eta$, explicitly
$$
\delta F=2\int_a^b[-(py')'+qy]\eta\,dx,\qquad\delta G=2\int_a^bwy\eta\,dx,
$$
so
$$
\delta\Lambda=\frac{\delta F-\Lambda\delta G}{G}=\frac2G\int_a^b[-(py')'+qy-\Lambda wy]\eta\,dx.
$$
The <fundamental lemma of the calculus of variations> shows that stationarity is exactly the <eigenvalue> equation with $\lambda=\Lambda$. Conversely an <eigenfunction> has vanishing first variation and quotient equal to its <eigenvalue>. Completeness gives the same weighted-average formula for any nonzero admissible expansion, even one with zero ground-state coefficient. Since every $\lambda_i\geq\lambda_0$,
$$
\boxed{\Lambda[y]\geq\lambda_0.}
$$
This is the <Rayleigh-Ritz variational principle>.
For the interval problem, the normalized <Dirichlet eigenfunctions> and their <eigenvalues>, with the paper's zero-based indexing, are
$$
y_i(x)=\sqrt2\sin[(i+1)\pi x],\qquad\lambda_i=(i+1)^2\pi^2.
$$
For $u=x(1-x)$, direct integration gives $F=\int_0^1(1-2x)^2dx=1/3$ and $G=\int_0^1x^2(1-x)^2dx=1/30$. Therefore
$$
\boxed{\lambda_{\rm trial}=10>\lambda_0=\pi^2.}
$$
The <sine expansion and Rayleigh quotient of a parabolic Dirichlet trial> follow from the coefficient integral
$$
\int_0^1x(1-x)y_i(x)dx=\frac{2\sqrt2[1-(-1)^{i+1}]}{(i+1)^3\pi^3}.
$$
Consequently the requested expansion has
$$
\boxed{A=\frac{4\sqrt2}{\pi^3},\qquad c_i=\frac{1-(-1)^{i+1}}{2(i+1)^3}.}
$$
Equivalently $u=(8/\pi^3)\sum_{n\geq1,\ n\text{ odd}}\sin(n\pi x)/n^3$. In particular $c_1=0$: the first excited <eigenfunction> $\sqrt2\sin(2\pi x)$ is antisymmetric about $1/2$, whereas the trial function is symmetric. Since $10-\pi^2>0$, the final ratio is exactly
$$
\boxed{\frac{c_1^2(\lambda_1-\lambda_0)}{\lambda_{\rm trial}-\lambda_0}=0.}
$$
The first nonzero correction is $c_2=1/27$, not $c_1$. Reindexing only the nonzero terms would change the meaning of the requested ratio and is not used here.
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