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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.

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  1. amistre64
    • one year ago
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    hmm, since all of our exponents are even, the results will always be postive for the first 3 terms ...

  2. amistre64
    • one year ago
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    i wonder if a gradient would help out any ...

  3. amistre64
    • one year ago
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    \[F(x,y,z)=2x^4y^2 + 9y^4z^2 + 12z^4x^2 - px^2y^2z^2\] \[F_x=8x^3y^2 + 24z^4x - 2pxy^2z^2\] \[F_y=4x^4y + 36y^3z^2 - 2px^2yz^2\] \[F_z=18y^4z + 48z^3x^2 - 2px^2y^2z\] wonder where i would go from there if this notion has any worth

  4. amistre64
    • one year ago
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    might 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 ...

  5. thomas5267
    • one year ago
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    It seems that p=0. Using Mathematica, the equation can be smaller than 0 even if p=0.0000000000000000001.

  6. thomas5267
    • one year ago
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    Alternatively, you can show that for all \(p>0\) there always exist \(x,y,z\) such that: \[ \frac{2 x^2}{p z^2} > 1\land \frac{9 y^2}{p x^2} > 1\land \frac{12 z^2}{p y^2}>1 \]

  7. thomas5267
    • one year ago
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    I did something wrong.

  8. zmudz
    • one year ago
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    @thomas5267 yea, p=0 isn't working :(

  9. thomas5267
    • one year ago
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    Showing 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.

  10. thomas5267
    • one year ago
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    The idea is to show that \(x^2y^2z^2\) dominates all other term for some \(x,y,z\).

  11. thomas5267
    • one year ago
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    I put the wrong function in Mathematica lol.

  12. thomas5267
    • one year ago
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    It seems like p=18.

  13. zmudz
    • one year ago
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    aha! Yay, so x^2 =3, y^3 = 2, x^2 = 1. that makes sense, and you can find it using am-gm. thanks for your help!

  14. thomas5267
    • one year ago
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    How to find it using AM-GM?

  15. zmudz
    • one year ago
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    when 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 \(2x^4y^2 = 9y^4z^2 = 12z^4x^2\)

  16. zmudz
    • one year ago
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    and that only works if \(x^2 = 3, y^2 = 2, z^2 = 1\) which gives p=18 once you plug it back in

  17. thomas5267
    • one year ago
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    Oh I see!

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