parmenides51 wrote: Given real numbers $x$ and $y$, such that $$x^4 y^2 + y^4 + 2 x^3 y + 6 x^2 y + x^2 + 8 \le 0 .$$Prove that $x \ge - \frac16$ Because \[x^4 y^2 + y^4 + 2 x^3 y + 6 x^2 y + x^2 + 8 = y^4+(x^2y+x+3)^2-6x-1.\]