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I want to maximize the function

$$f(x_1, x_2, x_3) = \dfrac{2\sqrt{3-(x_1+x_2+x_3)^2}}{x_2-x_1}$$

subject to the constraints

$$x_1^2 + x_2^2 + x_3^2 = 1$$ and $$0 \leq x_1 \leq x_3 \leq x_2 \leq 1.$$

So normally this function would be unbounded as $x_2 - x_1 \rightarrow 0$. But in this case, due to the second equality, we have that $(x_1, x_2, x_3) \rightarrow (x, x, x)$, and due to the fact that $(x_1, x_2, x_3)$ is on the positive octant of the unit sphere, we must have $x = \frac{1}{\sqrt{3}}$. In this case,

$$3 - (x_1 + x_2 + x_3)^2 \to 0$$

I want to show that this function is bounded by $2\sqrt{2}$, achieved by $(0, 1, 0)$, subject to these conditions, but I have no idea how to handle the inequalities. It's easily shown that

$$\sqrt{3-(x_1+x_2+x_3)^2} \leq \sqrt{2}$$

attained at $(0, 1, 0)$ under these conditions. You only have to note that you will not have blowup as $x_2 -x_1 \rightarrow 0$. However, I don't know how to show that this function does not go higher than $2\sqrt{2}$ as $x_2 - x_1 \rightarrow 0$.

Lagrange multipliers don't seem to help without computer assistance. Any advice in reasoning this out by hand is appreciated.

Rodrigo de Azevedo
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Wyatt Gregory
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  • I would say the objective function is strange, rather than the (linear and quadratic) constraints. – Rodrigo de Azevedo Jun 24 '20 at 19:22
  • You can express the second constraint as the intersection of three simpler ones, each of which can be quickly projected onto (which makes this more amenable to an algorithmic approach, which I know is not exactly what you're asking for): Up to reordering indices, you've got the intersection of (1) the monotone cone $x_1\leq\ldots\leq x_n$ and (2) $x_n\leq 1$ and (3) $x_1\geq 0$. (mono. cone projection avalable here http://proximity-operator.net/indicatorfunctions.html ) – Zim Jun 30 '20 at 22:13

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