Determining the minimum of a linear function subject to a quadratic inequality constraint What is the minimum value of $x+4z$, a function defined on $\mathbb{R^3}$, subject to the constraint $x^2 + y^2 +z^2 \leq 2$?
I know how to solve this if the constraint is an equality, but what shall I do if it's an inequality? Could anyone help me, please? 
 A: You might as well force $y=0$ since that gives the largest set of pairs $(x,z)$.

Thus, the new problem is to minimize $x+4z$, defined on $\{(x,z) \in \mathbb{R^2}\}$, subject to the constraint $x^2 + z^2 \leq 2$.

Clearly, the minimum value of $x+4z$ will be negative, and the point $(x,z)$ where the minimum occurs will have $x,z$ both non-positive (with at least one of$\;x,z\;$negative).

But then, for a point $(x,z)$ where the minimum occurs, scaling the point $(x,z)$ by a constant greater than $1$ would make $x+4z$ even more negative.

It follows that the minimum value of $x+4z$ will occur on the part of the circle $x^2 + z^2=2$ in the third quadrant (possibly including the endpoints of the arc).

So now you have minimization with an equality constraint.

Can you finish it?
A: The objective function $u=x+4z$ does not depend on $y$, therefore $y$ must be minimum, that is $y=0$ so that $x,z$ are the maximum negative values.
Thus the new optimization problem states: Minimize $u=x+4z$ subject to $x^2+z^2\le 2.$
The feasible region is a circle with the radius $\sqrt{2}$ and the contour line of the objective function is $z=-\frac14x+\frac14u$.
Minimum of $u(x,z)$ occurs when the contour line is tangent to the feasible region from below:
$$z=-\sqrt{2-x^2} \Rightarrow z'=\frac{x}{\sqrt{2-x^2}}=-\frac14 \Rightarrow x=-\sqrt{\frac{2}{17}} \Rightarrow z=-\sqrt{\frac{32}{17}} \Rightarrow$$
$$u=-\sqrt{\frac{2}{17}}-4\sqrt{\frac{32}{17}}=-\sqrt{34} \ (min).$$
A: We have the following quadratically constrained linear program (QCLP) in $\mathrm x \in \mathbb R^3$
$$\begin{array}{ll} \text{minimize} & \mathrm c^\top \mathrm x\\ \text{subject to} & \| \mathrm x \|_2^2 \leq 2\end{array}$$
where $\mathrm c := (1,0,4)$. Since the objective function is linear and nonzero, the minimum is attained at the boundary, i.e., on the Euclidean sphere of radius $\sqrt 2$ and centered at the origin. Hence, we have the following equality-constrained QCLP in $\mathrm x \in \mathbb R^3$
$$\begin{array}{ll} \text{minimize} & \mathrm c^\top \mathrm x\\ \text{subject to} & \| \mathrm x \|_2^2 = 2\end{array}$$

Via Cauchy-Schwarz
Using the Cauchy-Schwarz inequality,
$$\mathrm c^\top \mathrm x \geq - \| \mathrm c \|_2 \| \mathrm x \|_2 = - \sqrt{17} \cdot \sqrt{2} = - \sqrt{34}$$
Thus, the minimum is $\color{blue}{-\sqrt{34}}$, which is attained at $\mathrm x_{\min} = \gamma \mathrm c$, where $|\gamma| = \frac{\sqrt{2}}{\| \,\mathrm c \|_2} = \sqrt{\frac{2}{17}}$. Thus,
$$\mathrm x_{\min} = \color{blue}{- \begin{bmatrix} \sqrt{\frac{2}{17}}\\ 0\\ 4 \sqrt{\frac{2}{17}}\end{bmatrix}}$$

Via Lagrange multipliers
Let the Lagrangian be
$$\mathcal L (\mathrm x, \lambda) := \mathrm c^\top \mathrm x + \lambda \left( \| \mathrm x \|_2^2 - 2 \right)$$
Taking the partial derivatives of $\mathcal L$ and finding where they vanish, we obtain
$$\begin{array}{rl} \mathrm c + 2 \lambda \mathrm x &= 0_3\\ \| \mathrm x \|_2^2 &= 2\end{array}$$
Hence, $\mathrm x = - \frac{1}{2 \lambda} \mathrm c$ and, from $\| \mathrm x \|_2^2 = 2$, we obtain $\lambda = \pm \sqrt{\frac{17}{8}}$. Thus,
$$\mathrm x_{\min} = - \frac{1}{2 \lambda} \mathrm c = - \sqrt{\frac{2}{17}} \, \mathrm c = \color{blue}{- \begin{bmatrix} \sqrt{\frac{2}{17}}\\ 0\\ 4 \sqrt{\frac{2}{17}}\end{bmatrix}}$$
and the minimum is
$$\mathrm c^\top \mathrm x_{\min} = - \sqrt{\frac{2}{17}} \| \mathrm c \|_2^2 = \color{blue}{-\sqrt{34}}$$

Via SDP
Using the Schur complement test for positive semidefiniteness, the original inequality-constrained QCLP can be rewritten as the following semidefinite program (SDP)
$$\begin{array}{ll} \text{minimize} & \mathrm c^\top \mathrm x\\ \text{subject to} & \begin{bmatrix} \mathrm I_3 & \mathrm x \\ \mathrm x^{\top} & 2\end{bmatrix} \succeq \mathrm O_4\end{array}$$
