= Solution
The <Gram matrix> representation and $Y_{ii}=1$ give $\|v_i\|_2=1$. For a <standard Gaussian random vector> $Z$, each $\langle v_i,Z\rangle$ is a standard <normal> <scalar>, so the zero event has <probability> zero. Choose either sign convention at zero; the resulting <vector> is <almost surely> a feasible sign <vector>. Consequently $y^TAy\leq v^*$ <almost surely>.
The <Gaussian sign-correlation identity> gives
$$
\mathbb E[y^TAy]
=\sum_{i,j}A_{ij}\mathbb E[y_iy_j]
=\frac2\pi\sum_{i,j}A_{ij}\arcsin(Y_{ij})
=\frac2\pi\operatorname{tr}(A\arcsin[Y]).
$$
The last equality uses symmetry of the real <matrices>. Hence
$$
\boxed{v^*\geq\mathbb E[y^TAy]
=\frac2\pi\operatorname{tr}(A\arcsin[Y])}.
$$
The <inverse sine> is applied entrywise, not through <spectral matrix functional calculus>.
For completeness, the <correlation coefficient> identity has a geometric proof. If the angle between two unit <vectors> is $\theta$, the isotropic <Gaussian random vector> direction in their two-dimensional span gives opposite signs on angular sectors with <probability> $\theta/\pi$. Thus the sign product has <expectation> $1-2\theta/\pi=(2/\pi)\arcsin(\cos\theta)$. Parallel and antiparallel pairs give the endpoint values $1$ and $-1$ directly. This is <Gaussian hyperplane rounding>, and needs no <independence> between the rounded coordinates.
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