Question

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.

Answer 1

hmm, since all of our exponents are even, the results will always be postive for the first 3 terms ...

Answer 2

i wonder if a gradient would help out any ...

Answer 3

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

Answer 4

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 ...

Answer 5

It seems that p=0. Using Mathematica, the equation can be smaller than 0 even if p=0.0000000000000000001.

Answer 6

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

Answer 7

I did something wrong.

Answer 8

@thomas5267 yea, p=0 isn't working :(

Answer 9

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.

Answer 10

The idea is to show that \(x^2y^2z^2\) dominates all other term for some \(x,y,z\).

Answer 11

I put the wrong function in Mathematica lol.

Answer 12

It seems like p=18.

Answer 13

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!

Answer 14

How to find it using AM-GM?

Answer 15

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

\(2x^4y^2 = 9y^4z^2 = 12z^4x^2\)

Answer 16

and that only works if \(x^2 = 3, y^2 = 2, z^2 = 1\) which gives p=18 once you plug it back in

Answer 17

Oh I see!