quasi wrote: > Peter Percival wrote: >> quasi wrote:
>>> >>> No, that's not automatic -- it requires proof. >> >> Indeed so. What I meant when I said it was clear is that the >> proof is very simple and can be be done in ones head. >> >> For simplicity's sake let n = 2, x_1 = y and x_2 = z, then my >> claim is that if >> >> (1 + y^2 + z^2)/(1 + y + z) >> >> has a minimum, then y = z there. Why is that clear? Because >> if it weren't so then >> >> (1 + y^2 + z^2)/(1 + y + z) =/= (1 + z^2 + y^2)/(1 + z + y). > > No, your appeal to symmetry doesn't support your claim unless > you know that there is a _unique_ point in R^2 for which the > function achieves a minimum.
Silly me! What a foolish mistake to make.
> If all you know is that the function f is symmetric in the > variables y and z and that f achieves a global minimum value > somewhere in the domain, that's not enough to conclude that > the global minimum value can be achieved at some point for > which y = z. > > quasi >
-- Sorrow in all lands, and grievous omens. Great anger in the dragon of the hills, And silent now the earth's green oracles That will not speak again of innocence. David Sutton -- Geomancies