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my problem (P) is:
$$(P) \space \space \text{min} \space x_1x_2$$ $$\text{s.t.} \space x_1-x_2-2 \leq 0$$ $$x_2 \leq 0$$
Prove that $x^* = (1,-1)$ is a strict local minimizer.
Since there are only inequality constraints, what method would I use to check this is a local minimizer ?
You can show this directly.
Multiplying the constraints gives $x_1 x_2 \ge x_2(x_2+2)$. Since $\min_x x(x+2) = -1$, we have $x_1 x_2 \ge -1$. Furthermore, since $x(x+2) = -1$ iff $x=-1$, we see that, in fact, $x_1 x_2 > -1$ for $x_2 \neq -1$.
Evaluating the cost at $(1,-1)$ gives $-1$, so we see that $(1,-1)$ is a global minimiser. We have already shown that this is strict when $x_2 \neq -1$, so now consider $x_2 = -1$ in which case the first constraint gives $x_1 \le 1$. If $x_1 < 1$, then $x_1 x_2 = - x_1 > -1$, from which it follows that $(1,-1)$ is a strict global minimiser.
