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zmudz
 one year ago
Find the maximum \(p\) such that \(2x^4y^2 + 9y^4z^2 + 12z^4x^2  px^2y^2z^2\) is always nonnegative for all \(x\), \(y\), and \(z\) real.
zmudz
 one year ago
Find the maximum \(p\) such that \(2x^4y^2 + 9y^4z^2 + 12z^4x^2  px^2y^2z^2\) is always nonnegative for all \(x\), \(y\), and \(z\) real.

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amistre64
 one year ago
Best ResponseYou've already chosen the best response.0hmm, since all of our exponents are even, the results will always be postive for the first 3 terms ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i wonder if a gradient would help out any ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0\[F(x,y,z)=2x^4y^2 + 9y^4z^2 + 12z^4x^2  px^2y^2z^2\] \[F_x=8x^3y^2 + 24z^4x  2pxy^2z^2\] \[F_y=4x^4y + 36y^3z^2  2px^2yz^2\] \[F_z=18y^4z + 48z^3x^2  2px^2y^2z\] wonder where i would go from there if this notion has any worth

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0might want to work this for p instead of F 2 x^4 y^2 + 9 y^4 z^2 + 12 z^4 x^2  p x^2 y^2 z^2 >= 0 2 x^4 y^2 + 9 y^4 z^2 + 12 z^4 x^2 >= p x^2 y^2 z^2 2 x^4 y^2 + 9 y^4 z^2 + 12 z^4 x^2  >= p x^2 y^2 z^2 2 x^2 + 9 y^2 + 12 z^2    >= p z^2 x^2 y^2 then my idea of min/max might be useful, but then again it might not ...

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1It seems that p=0. Using Mathematica, the equation can be smaller than 0 even if p=0.0000000000000000001.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Alternatively, you can show that for all \(p>0\) there always exist \(x,y,z\) such that: \[ \frac{2 x^2}{p z^2} > 1\land \frac{9 y^2}{p x^2} > 1\land \frac{12 z^2}{p y^2}>1 \]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1I did something wrong.

zmudz
 one year ago
Best ResponseYou've already chosen the best response.1@thomas5267 yea, p=0 isn't working :(

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Showing that for all \(p>0\) there exists \(x,y,z\) such that: \[ \frac{p z^2}{2 x^2} > 1\land \frac{p x^2}{9 y^2} > 1\land \frac{p y^2}{12 z^2}>1 \] will show that p=0 is the maximum p.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1The idea is to show that \(x^2y^2z^2\) dominates all other term for some \(x,y,z\).

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1I put the wrong function in Mathematica lol.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1It seems like p=18.

zmudz
 one year ago
Best ResponseYou've already chosen the best response.1aha! Yay, so x^2 =3, y^3 = 2, x^2 = 1. that makes sense, and you can find it using amgm. thanks for your help!

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1How to find it using AMGM?

zmudz
 one year ago
Best ResponseYou've already chosen the best response.1when you apply the inequality, you should get something like \(\frac{2x^4y^2 + 9y^4z^2 + 12z^4x^2}{3} \geq 6x^2y^2z^2\) this simplifies to \(2x^4y^2 + 9y^4z^2 + 12z^4x^2  18x^2y^2z^2 \geq 0\) then that only works if \(2x^4y^2 = 9y^4z^2 = 12z^4x^2\)

zmudz
 one year ago
Best ResponseYou've already chosen the best response.1and that only works if \(x^2 = 3, y^2 = 2, z^2 = 1\) which gives p=18 once you plug it back in
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