Find the min value of $\min\{x_1x_2x_3:a^2x_1^2+x_2^2+x_3^2\leq1\},a>0$

Question

Find the min value of $\min\{x_1x_2x_3:a^2x_1^2+x_2^2+x_3^2\leq1\},a>0$ using KKT
So my try is:
if we set $L(x,\lambda)=x_1x_2x_3+\lambda(a^2x_1^2+x_2^2+x_3^2-1)$ and differentiating wrt to all variables we get:
$\bullet$$x_2x_3+2\lambda a^2x_1=0$
$\bullet$$x_1x_3+2\lambda x_2=0$
$\bullet$$x_1x_2+2\lambda x_3=0$
$\bullet$$\lambda(a^2x_1^2+x_2^2+x_3^2-1)$
I concluded that all $x_i\neq 0$ and $\lambda>0$ then to every equallity $i=1,2,3$ we multiply by $x_i$ and then sum all the first 3 equations we get that $x_1x_2x_3=\frac{-2\lambda}{3}$ but I got stuck here and didn't know what to do next to find the minimum value for this porblem

Answer 1

From the Lagrangian above, the FONCs are \begin{eqnarray*} x_2 x_3 + 2 a^2 x_1 &=& 0\\ x_1 x_3 + 2 \lambda x_2 &=& 0\\ x_1 x_2 + 2 \lambda x_3 = 0 \end{eqnarray*} and the constraint $a^2 x_1^2 + x_2^2 + x_3^2 \le 1$.

Let's take a second to think about this. We want to minimize, so to get the lowest number possible. The objective will be negative as long as one or three of the controls are negative. If any of the controls are zero, the objective is zero, and I can do better by readjusting all of the controls slightly negative so that the constraint binds. Since a negative value is achievable, the constraint will bind, since there will be a feasible direction in which we can continue decreasing the function. But that means there are probably many solutions to be careful about.

Since we're not worried about zero controls, solving the FONCs in terms of $\lambda$ and setting them all equal yields $$ \dfrac{x_2 x_3}{ 2a^2 x_1} = \dfrac{ x_1 x_3}{2 x_2} = \dfrac{ x_1 x_2}{ 2x_3}, $$ The first and second terms yield $x_1^2 a^2=x_2^2$, and the second and third yield $x_3^2 = x_2^2$, and the first and third yield $x_3^2 = a^2 x_1^2$.

Since $\sqrt{x^2} = |x|$, this implies $|x_1^*|a=|x_2^*|=|x_3^*|$ at any solution. Substituting the terms at the end of the last paragraph into the constraint yields $3 x_2^2 =1$, so $|x_2^*| = 1/\sqrt{3}$; $3 x_3^2 = 1$, so $|x_3^*|=1/\sqrt{3}$; and $3 a^2 x_1^2 = 1$, so $|x_1^*| = 1/\sqrt{3a}$.

So any solution takes the following form: either one or three of the controls are negative, and $|x_1^*|=1/\sqrt{3a}$, $|x_2^*|=1/\sqrt{3}$, and $|x_3^*|=1/\sqrt{3}$. That's 4 possible solutions in total.