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Re: inequality (a quickie)

Posted: September 22nd, 2019, 7:32 pm
by Cuchulainn
 We are looking for a minimum (of a positive function).
Not any old function. The original is 

[$]\sqrt[3]{\frac{x^3+y^3+z^3} {xyz}}+  \sqrt[2]{\frac{xy+yz+zx} {x^2 + y^2 +  z^2}} \geq \sqrt[3]{3} + 1 [$]

which is not 

 [$] f(x, y, z) = \frac{x^2 + y^2 + z^2}{x y + y z + z x} [$]

The function has been inverted. Why? The problem has been changed.

Re: inequality (a quickie)

Posted: September 22nd, 2019, 8:00 pm
by Paul
It would be a silly question if the first term on the left was always greater than the first on the right and ditto for the second term.

Re: inequality (a quickie)

Posted: September 23rd, 2019, 5:40 am
by Cuchulainn
It would be a silly question if the first term on the left was always greater than the first on the right and ditto for the second term.
Exactly. I think it was concocted by working backwards from the solution. It has further probably little saving graces.

P.S. what I am saying is that these puzzles should relate to something like geometry etc.

Re: inequality (a quickie)

Posted: September 24th, 2019, 6:05 pm
by katastrofa
It would be a silly question if the first term on the left was always greater than the first on the right and ditto for the second term.
Exactly. I think it was concocted by working backwards from the solution. It has further probably little saving graces.

P.S. what I am saying is that these puzzles should relate to something like geometry etc.
Like hyperspheres and hyperboloids? I can only imagine the two terms separately, and the same can be easily shown analytically:
[$]\sqrt[3]{\frac{x^3+y^3+z^3} {xyz}}\geq \sqrt[3]{3}[$] because [$]\frac{x^3+y^3+z^3}{3} \geq \sqrt[3]{x^3y^3z^3}[$], but
[$]\sqrt[2]{\frac{xy+yz+zx} {x^2 + y^2 +  z^2}} \leq 1[$] because [$](x_i-x_j)^2 >0\Rightarrow xy+yz+zx\leq x^2 + y^2 +  z^2[$], obviously.