I'm trying to solve this problem by KKT's condition:
$$\begin{align*} \text{min} & \quad x + z \\ \text{s.t} & \quad x^2 + y^2 = 1 \\ & \quad y^2+z^2 = 4 \end{align*}$$
One of the regularity conditions is linear independence constraint constraint qualification (LICQ), which is
The gradients of the active inequality constraints and the gradients of the equality constraints are linearly independent at $x^{*}$.
Could you please verify if I correctly apply the KKT's theorem? Thank you so much for your help!
$\textbf{My attempt:}$
Put $f(x,y,z) = x + z$, $h_1(x,y,z) = x^2+y^2-1$, $h_2(x,y,z)=y^2 + z^2 -4$, and $\mathcal K= \{(x,y,z) \in \mathbb R^3 \mid h_1(x,y,z) = h_2(x,y,z) = 0\}$.
Because $\mathcal K$ is compact and $f$ is continuous, the problem has a solution.
Moreover, $\nabla f (x,y,z) = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$, $\nabla h_1 (x,y,z) = \begin{pmatrix} 2x \\ 2y \\ 0 \end{pmatrix}$, and $\nabla h_2 (x,y,z) = \begin{pmatrix} 0\\ 2y \\ 2z \\ \end{pmatrix}$.
Next, we check if LICQ is satisfied by considering the system of equations: $$\begin{cases} \mu_1 \nabla h_1 (x,y,z) + \mu_2 \nabla h_2 (x,y,z)=0 &=0 \\ h_1(x,y,z) &= 0\\ h_1(x,y,z) &=0 \end{cases} \iff \begin{cases} \mu_1 \begin{pmatrix} 2x \\ 2y \\ 0 \end{pmatrix} + \mu_2 \begin{pmatrix} 0\\ 2y \\ 2z \\ \end{pmatrix} &= \begin{pmatrix} 0\\ 0 \\ 0 \\ \end{pmatrix} \\ x^2 + y^2 &= 1\\ y^2+z^2 &= 4 \end{cases}$$
$$\iff \begin{cases} x \mu_1 &= 0\\ y(\mu_1 + \mu_2) &= 0 \\ z \mu_2 &= 0 \\ x^2 + y^2 &= 1\\ y^2+z^2 &= 4 \end{cases}$$
If $\mu_1 \neq 0$ then $x=0$. It follows that $y^2 =1$ and $z^2 =3$. Hence $\mu_2=0$ and thus $y(\mu_1 + \mu_2) \neq 0$, which is impossible. If $\mu_2 \neq 0$ then $z=0$. It follows that $y^2 =4 >1$, which is impossible. As such, $\mu_1= \mu_2 =0$ and thus LICQ is satisfied. It follows from KKT's theorem that the solution satisfies
$$\begin{cases} h_1(x,y,z) &= 0\\ h_1(x,y,z) &=0 \\ \nabla f (x,y,z) + \mu_{1} \nabla h_1(x,y,z) + \mu_{2} \nabla h_1(x,y,z)&=0 \end{cases} \iff \begin{cases} x^2 + y^2 &= 1\\ y^2+z^2 &= 4 \\ \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} +\mu_1 \begin{pmatrix} 2x \\ 2y \\ 0 \end{pmatrix} + \mu_2 \begin{pmatrix} 0\\ 2y \\ 2z \\ \end{pmatrix} &= \begin{pmatrix} 0\\ 0 \\ 0 \\ \end{pmatrix} \end{cases}$$ $$\iff \begin{cases} x^2 + y^2 &= 1\\ y^2+z^2 &= 4 \\ 2 \mu_1 x &= -1\\ y(\mu_1 + \mu_2) &=0 \\ 2 \mu_2 z &= -1 \end{cases} \iff \begin{cases} x^2 &= 1\\ z^2 &= 4 \\ y &= 0\\ \end{cases}$$
Hence $(x,y,z) =(\pm1,0,\pm2)$. After comparing the values at these points, we get the solution is $(-1,0,-2)$.
Yes, your solution is correct, I just corrected a small typo in the notations of $f(x, y, z)$.