A Beautiful but Difficult Inequality

@Daniel Liu – The first condition that needs to be met is that the functions involved must have continuous first partial derivatives. This is the case here as long as at most one of the variables is equal to $0$ , (which must be the case for the given function to be continuos anyway). Applying Lagrange under this condition yields the symmetry solution Now while no alarm bells go off if we were to just set one of the variables equal to $0$ , when we do set, say, $z = 0$ and then apply Lagrange, we end up with the given function being maximized for $(x,y) = (3,1)$ . So I guess we might have to consider the $z = 0$ scenario as an "endpoint" that we need to check given the non-negative requirement, but I can't yet make a formal argument for this.

Note that if we have $(x,y,z) = (3,1 - \epsilon, \epsilon)$ then $\lim_{\epsilon \rightarrow 0+} f(x,y,z) = 2.5$ , so the maximum is not some isolated point. In fact, this limit would appear to be two-sided, all the more reason to have expected the Lagrange method to flag it as a local maximum, and yet this didn't happen.

@Pi Han Goh – Yeap maybe tomorow...It's late.