Complicated non-linear inequality reduces to simple linear equality. How?

Question

I'm solving a complex inequality, and I obtain the expression $$8xy+x^2y+2\ge3x^3+5x^2+3xy^2+4x-y^3+y^2$$ and I get stuck. This is somehow equivalent to $y\ge3x-1$ which means that I got it correct however I'm unable to show that the expression I obtain reduces to the simple linear inequality. My guess is that there is probably a neat factorisation of the terms leading to cancelation but I am unable to find it.

Answer 1

Yes, you are right: $$8xy+x^2y+2-(3x^3+5x^2+3xy^2+4x-y^3+y^2)=$$ $$=(y+1-3x)((x+1)^2+(y-1)^2)\geq0.$$

We can rewrite our inequality in the following form: $$y^3-(3x+1)y^2+(x^2+8x)y-3x^3-5x^2-4x+2\geq0.$$ Now, easy to see that $\frac{1}{3}$ is a root of $3x^3+5x^2+4x-2,$ which gives: $$3x^3+5x^2+4x-2=3x^3-x^2+6x^2-2x+6x-2=(3x-1)(x^2+2x+2).$$ Id est, we need to prove that $$y^3-(3x+1)y^2+(x^2+8x)y-(3x-1)(x^2+2x+2)\geq0.$$ Now, we can check a factor $y-3x+1$.

Indeed, $$y^3-(3x+1)y^2+(x^2+8x)y-(3x-1)(x^2+2x+2)=$$ $$=y^3-(3x+1)y^2+6xy-2y+(x^2+2x+2)y-(3x-1)(x^2+2x+2)=$$ $$=y^3-(3x-1)y^2-2y^2+6xy-2y+(x^2+2x+2)y-(3x-1)(x^2+2x+2)=$$ $$=(y-(3x-1))(y^2-2y+x^2+2x+2)=(y-3x+1)((x+1)^2+(y-1)^2)\geq0.$$

Answer 2

Perhaps if you let $y=3x-1+z$, then $z$ might emerge as a factor, leaving a quadratic that you might be able to show is positive.