<s>[INST] Why are the solutions of the $3\times 3$ system like that? Consider the problem:$$
\min \quad -x_1^2-4x_1x_2-x_2^2\\ \text{s.t.} \quad x_1^2 + x_2^2 = 1$$
The KKT system is given by\begin{align*}
x_1 (-1 + v) + 2 x_2 &= 0 \tag{1} ,\\ x_2 (-1 + v) + 2 x_1 &= 0 \tag{2},\\ x_1^2 + x_2^2 &= 1 \tag{3}
\end{align*}
The solutions according to the book are$$
(x_1,x_2)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right),\ v=3;\\
(x_1,x_2)=\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right),\ v=3;\\
(x_1,x_2)=\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right),\ v=-1,\\
(x_1,x_2)=\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right),\ v=-1,\\
$$
I tried to solved by hand: Substracting (1) and (2) I get $(x_1-x_2)(1+v)=0$. As $v$ cannot be zero (by the KKT condition) we have that $x_1=x_2$ if $v\neq -1$.
Thus I get the solutions:$$
(x_1,x_2)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right);\\
(x_1,x_2)=\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right);\\
(x_1,x_2)=\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right);\\
(x_1,x_2)=\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right);\\
$$
only asking $v\neq -1$.
Why the book ask $v=-1$ and $v=3$? Could someone explain please?
 [/INST] $(x_1-x_2)^2\geq0$ gives $$x_1x_2\leq\frac{x_1^2+x_2^2}{2}=\frac{1}{2}.$$
The equality occurs for $x_1=x_2$.
Thus,
$$-x_1^2-4x_1x_2-x_2^2=-1-4x_1x_2\geq-1-4\cdot\frac{1}{2}=-3.$$
The equality occurs for
$$(x_1,x_2)\in\left\{\left(\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right),\left(-\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\right)\right\}$$ only.
</s>