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

Completing the square we have $$(x^2y)^2+2(x^2y)x+x^2+2(x^2y)3+3^2+6x\le(1+6x)$$
$$\iff1+6x\ge(x^2y+x+3)^2$$ which is $\ge0$ for real $x,y$
