= Solution
The two <algebras> have the same identity, so an inverse in $A$ is also an inverse in $B$. Therefore \b[$G(A)\subseteq G_B(A)$].
For strictness, take $B=C_{\mathbb C}(\mathbb T)$, where $\mathbb T$ is the unit circle, and let $A$ be the uniform closure of the <polynomials> in the coordinate function $z$. This is a closed unital subalgebra of the <Banach algebra> $B$. The coordinate $z$ has inverse $\overline z$ in $B$. But $\overline z$ cannot belong to $A$: for every <polynomial> $p$,
$$
\frac1{2\pi}\int_0^{2\pi} e^{it}p(e^{it})\,dt=0,
$$
and the same identity holds for every element of $A$ by <uniform convergence>. It fails for $\overline z$, since $z\overline z=1$. Thus \b[$z\in G_B(A)\setminus G(A)$]. This <algebra> is the boundary realization of the <disk algebra>.
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