<s>[INST] Show that if x,y are and $ x^4y^2+x^2+2x^3y+6x^2y+8 \leq 0 $ then $x \geq -1/6 $ Show that if x,y are real and $ x^4y^2+x^2+2x^3y+6x^2y+8 \leq 0 $ then $x \geq -1/6 $
So far I've tried factoring $x^2$ and throwing the 8 on the LHS, but can't get to the needed result. Help would be appreciated
 [/INST] Maybe it's not the clever way, but it works...
You have $x^4y^2 + (2x^3 + 6x^2)y + x^2 + 8 \leq 0$ for some $y$
Then you should have $(2x^3 + 6x^2)^2 - 4 x^4(x^2 + 8) \geq 0$
this is to say $24x^5 + 4x^4 \geq 0$
$x$ can be $0$ of course, and when $x \neq 0$, divide both sides by $x^4$ we get $24x + 4 \geq 0$, i.e. $x\geq -\frac{1}{6}$
</s>