= Solution
The <Lagrangian sufficiency theorem> for a maximization problem says the following. Suppose the constraints are $g_i(q)\ge0$ and $h_j(q)=0$, and define
$$
L(q,\lambda,\mu)
=f(q)+\sum_i\lambda_i g_i(q)+\sum_j\mu_jh_j(q),
\qquad \lambda_i\ge0.
$$
If a feasible $q^*$ satisfies <complementary slackness> $\lambda_i g_i(q^*)=0$ and globally maximizes $L(\,\cdot\,,\lambda,\mu)$, then $q^*$ globally maximizes $f$. Indeed, for every feasible $q$,
$$
f(q)\le L(q,\lambda,\mu)
\le L(q^*,\lambda,\mu)
=f(q^*).
$$
This proves the theorem. In a <concave function>[concave maximization problem], stationarity and the boundary optimality conditions ensure the required global maximum of the Lagrangian.
For the problem at hand, retain $x,z\ge0$ as the domain and attach a multiplier $\lambda$ to the equality:
$$
L=x+y+2a\sqrt{1+z}
+\lambda\left(b-x-\frac12y^2-z\right).
$$
For $\lambda\ge1$, this is concave in $(x,y,z)$ on the domain. Its separate maximizers satisfy
$$
y=\frac1\lambda,
\qquad
z=\max\left(\frac{a^2}{\lambda^2}-1,0\right),
$$
while the $x$ term is $(1-\lambda)x$. Thus $x$ may be positive only when $\lambda=1$; when $\lambda>1$, its maximizing value is $x=0$.
If
$$
b\ge a^2-\frac12,
$$
take $\lambda=1$. Then $y=1$, $z=a^2-1$, and feasibility fixes
$$
x=b-a^2+\frac12\ge0.
$$
The Lagrangian sufficiency theorem gives the maximum
$$
\boxed{(x,y,z)=
\left(b-a^2+\frac12,\,1,\,a^2-1\right)},
$$
$$
\boxed{f_{\max}=b+a^2+\frac32}.
$$
If
$$
\frac12\le b<a^2-\frac12,
$$
then $\lambda>1$ and $x=0$. The equality constraint becomes
$$
\frac1{2\lambda^2}
+\frac{a^2}{\lambda^2}-1=b,
$$
so
$$
\lambda^2=\frac{a^2+1/2}{b+1}.
$$
Consequently
$$
\boxed{x=0,\qquad
y=\sqrt{\frac{2(b+1)}{2a^2+1}},
\qquad
z=\frac{2a^2b-1}{2a^2+1}}.
$$
The assumptions $a\ge1$ and $b\ge1/2$ ensure $z\ge0$. At the stationary point $\sqrt{1+z}=ay$, so the maximum value is
$$
\boxed{f_{\max}
=(1+2a^2)y
=\sqrt{2(b+1)(2a^2+1)}}.
$$
The two formulas agree at $b=a^2-1/2$.
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