MHB Proving expression is always positive

[Carla1985]

I am trying to prove that

$$
x^4+4096y^4+450x^2y^2-2304xy^3-36x^3y
$$

is positive for all positive $x$ and $y$. I also have the condition $x>16y$ though I don't believe this is needed to prove positivity as I've plotted the function for varying $x$ and $y$ and it always seems to be positive. I've tried completing the square etc in a variety of ways but I always seem to be a term out of getting them all positive. I think the closest I've got is

$$
(x^2-18xy+64y^2)^2-2x^2y^2
$$

Could someone look at it please. I think I've been looking at it for too long and am stuck in the methods I'm using.

Thanks
Carla

 

[Opalg]

I am trying to prove that

$$
x^4+4096y^4+450x^2y^2-2304xy^3-36x^3y
$$

is positive for all positive $x$ and $y$. I also have the condition $x>16y$ though I don't believe this is needed to prove positivity as I've plotted the function for varying $x$ and $y$ and it always seems to be positive. I've tried completing the square etc in a variety of ways but I always seem to be a term out of getting them all positive. I think the closest I've got is

$$
(x^2-18xy+64y^2)^2-2x^2y^2
$$

Could someone look at it please. I think I've been looking at it for too long and am stuck in the methods I'm using.

Thanks
Carla

If $x=5$ and $y=1$ then $(x^2-18xy+64y^2)^2-2x^2y^2 = (25 - 90 + 64)^2 - 50 = 1-50 = -49 <0,$ so the condition $x > 16y$ is certainly needed.

I think that your expression $(x^2-18xy+64y^2)^2-2x^2y^2$ is a very good start. If we write $t = \dfrac xy$ then we can write it as $y^4f(t)$, where $f(t) = (t^2 - 18t + 64)^2 - 2t^2.$ Factorise that as the difference of two squares: $$f(t) = \bigl(t^2 - (18+\sqrt2)t +64\bigr) \bigl(t^2 - (18-\sqrt2)t +64\bigr).$$ You need to show that both factors are positive when $t>16$.

To do that, I would look at the more general quadratic expression $t^2 - \lambda t + 64$. That expression will always be positive when $t$ is greater than the larger of its two roots, namely $\frac12\bigl(\lambda + \sqrt{\lambda^2 - 256}\bigr).$ You want that root to be less than $16$, and you can check that this will be the case when $\lambda<20$. The numbers $18\pm\sqrt2$ are both less than $20$, so everything fits together nicely!

 

[Carla1985]

That's fab, thank you so much.

I'll work though that method. Again, thank you.