= Solution
Convert to minimization with $f=-F$ and retain $g\geq0$. A continuously differentiable inequality <augmented Lagrangian method> uses the <merit function>
$$
\boxed{\Phi(x,\lambda,\mu)=f(x)+\frac1{2\mu}\sum_{i=1}^3
\left(\min\{g_i(x)-\mu\lambda_i,0\}^2-\mu^2\lambda_i^2\right),
\quad\mu>0,\quad\lambda\geq0.}
$$
For $g_i\leq\mu\lambda_i$ its contribution is $-\lambda_i g_i+g_i^2/(2\mu)$; otherwise it is the constant $-\mu\lambda_i^2/2$. This shifted truncation permits multipliers to decrease as well as increase while remaining nonnegative.
A useful initial estimate comes from the preceding pure <quadratic penalty method> at $\mu_{-1}$:
$$
\lambda_i^{[0]}=\max\{0,-g_i(x^{[0]})/\mu_{-1}\}.
$$
Here $g(-1,-1)=(3,18,-2)$, so with $\mu_{-1}=2$ choose
$$
\boxed{\lambda^{[0]}=(0,0,1).}
$$
Then use this estimate and $\mu_0=1$ in the first augmented subproblem. Initial estimates are a choice, rather than uniquely specified data; zero multipliers would also define an initialization.
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