parmenides51 wrote: For non negative reals $a,b$ we know that $a^2+a+b^2\ge a^4+a^3+b^4$. Prove that $$\frac{1-a^4}{a^2}\ge \frac{b^2-1}{b}$$ $(a-1)^2(a^2+3a+2)+(b-1)^2(b^2+2b+2)\geq0$ so $4-2a^2-2b-((a^2+a+b^2)-( a^4+a^3+b^4))\ge 0$. Then $a^2b\le a^2(2-a^2)\le 1$ so $b-\frac{1}{b}\le \frac{1}{a^2}-a^2$.