## 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.

1. amistre64

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

2. amistre64

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

3. amistre64

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

4. amistre64

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

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

6. thomas5267

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

I did something wrong.

8. zmudz

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

9. thomas5267

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

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

11. thomas5267

I put the wrong function in Mathematica lol.

12. thomas5267

It seems like p=18.

13. zmudz

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

How to find it using AM-GM?

15. zmudz

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

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

Oh I see!