= Solution
Put $t=\log(1+\sqrt2)=\operatorname{arsinh}(1)$, so $c_K=2t/\pi$. The <power series> <coefficients> of the <hyperbolic sine> preprocessing are
$$
f_{2k}=g_{2k}=0,\qquad
f_{2k+1}=\frac{t^{2k+1}}{(2k+1)!},\qquad
g_{2k+1}=(-1)^kf_{2k+1}.
$$
Thus $f_j=|g_j|$ for every $j\geq0$, and the series converge on the full interval. Applying <coefficient-dominated entrywise positivity> gives $Y\succeq0$.
Every diagonal entry belongs to one of the diagonal blocks and equals $f(X_{ii})=f(1)=\sinh t$. Because $e^t=1+\sqrt2$ and $e^{-t}=\sqrt2-1$, this is $(e^t-e^{-t})/2=1$. Therefore
$$
\boxed{Y\succeq0,\qquad Y_{ii}=1}.
$$
The <matrix> lies in the <elliptope> and admits a <Gram matrix> representation by unit <vectors>. This preprocessing is the <Krivine rounding scheme>; the equality of absolute <coefficients> is what preserves <positive semidefiniteness> even though the cross-block <sine> <coefficients> alternate in sign.
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