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• 2 Cuchulainn
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### Re: inequality (a quickie)

We are looking for a minimum (of a positive function).
Not any old function. The original is

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

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.
Step over the gap, not into it. Watch the space between platform and train.
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http://www.datasim.nl Paul
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Joined: July 20th, 2001, 3:28 pm

### Re: inequality (a quickie)

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. Cuchulainn
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Joined: July 16th, 2004, 7:38 am
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### Re: inequality (a quickie)

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.
Step over the gap, not into it. Watch the space between platform and train.
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http://www.datasim.nl katastrofa
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Joined: August 16th, 2007, 5:36 am
Location: Alpha Centauri

### Re: inequality (a quickie)

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:
$\sqrt{\frac{x^3+y^3+z^3} {xyz}}\geq \sqrt{3}$ because $\frac{x^3+y^3+z^3}{3} \geq \sqrt{x^3y^3z^3}$, but
$\sqrt{\frac{xy+yz+zx} {x^2 + y^2 + z^2}} \leq 1$ because $(x_i-x_j)^2 >0\Rightarrow xy+yz+zx\leq x^2 + y^2 + z^2$, obviously.  