Prove that $a^3b^3 + b^3 + 1 \geq a^2b^2 + ab^3 + b$ (for real, positive $a$,$b$

Question

I was working on an Olympiad-level inequality, which I was able to boil down to the following inequality:

Prove that: $a^3b^3 + b^3 + 1 \geq a^2b^2 + ab^3 + b$

I think it's more useful to write it as:

Prove that: $a^3b^3 + b^3 + 1 - a^2b^2 - ab^3 - b \geq 0$

I'm pretty sure that this is true, simply by checking values for $a$ and $b$ via sliders on Desmos. Normally, to prove an inequality of this type, I would try to rewrite it as the sum of smaller, more manageable terms, which individually would all be squares, leading to a greater than $0$ sum. But here I struggled to do the same - for example, $a^3b^3 - a^2b^2 - ab^3 $ is not always greater than 0.

I realize that, in all likelihood, there probably exists a nice way to prove the original inequality I wanted to show without having to prove this one. However, this phenomenon occurs rather often for me - I boil down a complex inequality into a correct, simpler one, but cannot actually prove it, thus failing at the last hurdle. Is there a way to brute force these small inequalities and show them to be true, ideally with pre-calculus mathematics?

Answer 1

Let $ab=x$, $b=y$ and $1=z$.

Thus, we need to prove that: $$x^3+y^3+z^3\geq x^2z+y^2x+z^2y,$$ which is true by Rearrangement.

The last inequality we can prove also by SOS: $$x^3+y^3+z^3-(x^2z+y^2x+z^2y)=\sum_{cyc}(x^3-xy^2)=$$ $$=\sum_{cyc}\left((x-y)(x^2+xy)-\frac{2}{3}(x^3-y^3)\right)=\frac{1}{3}\sum_{cyc}(x-y)^2(x+2y)\geq0.$$

Answer 2

Thanks to Calvin Lin, and his suggestion to just keep trying to use the AM-GM inequality.

I multiplied both sides by $3$, at which point:

$$a^3b^3+b^3+b^3 \geq 3(a^3b^9)^{\frac{1}{3}} = 3ab^3$$ $$a^3b^3+a^3b^3+1 \geq 3(a^6b^6)^{\frac{1}{3}} = 3a^2b^2$$ $$b^3+1+1 \geq 3(b^3)^{\frac{1}{3}} = 3b $$

Adding together the left and right hand sides of all three equations and then dividing by $3$ gives the original inequality.