If the formulas $x^3-3y^2x \ge 3x^2y-y^3$ and $x+y=-1$ together define a line segment, find the length of this line segment.
I've tried setting $x=-(1+y)$, substituting that into the first equation and solving it when equality holds, and then finding the length of the line segment which I got as $\sqrt{30}/6$. This is very messy and I'm sure that there is a better approach.
On
The segment $\sigma$ in question is formed the points on the line $\ell:\ x+y=-1$ satisfying $$(x+y)^3-6xy(x+y)\geq0\ ,$$ which is the same thing as the points on $\ell$ satisfying $6xy\geq1$. For these points we also have $$(x-y)^2=(x+y)^2-4xy\leq1-{2\over3}\ ,$$ hence $|x-y|\leq{1\over\sqrt{3}}$. The two lines $y=x\pm{1\over\sqrt{3}}$ intersect $\ell$ orthogonally, and are at distance ${1\over\sqrt{2}}\cdot{2\over\sqrt{3}}={\sqrt{6}\over3}$ of each other. It follows that this distance is then also the length of $\sigma$.
Note that $$\begin{align}&x^3-3y^2x\ge 3x^2y-y^3\\\\&\iff (x^3+y^3)-3xy(x+y)\ge 0\\\\&\iff (x+y)(x^2-xy+y^2)-3xy(x+y)\ge 0\\\\&\iff (x+y)((x+y)^2-3xy)-3xy(x+y)\ge 0\end{align}$$
Since $x+y=-1$, we get $$(-1)((-1)^2-3xy)-3xy(-1)\ge 0,$$ i.e. $$xy\ge \frac 16,$$ i.e. $$x(-x-1)\ge\frac 16,$$ i.e. $$(x_{\text{min}}=)\ \frac{-3-\sqrt 3}{6}\le x\le \frac{-3+\sqrt 3}{6}\ (=x_{\text{max}})$$
Therefore, the answer is $$\sqrt{(x_{\text{max}}-x_{\text{min}})^2+((-x_{\text{max}}-1)-(-x_{\text{min}}-1))^2}=\sqrt 2\ |x_{\text{max}}-x_{\text{min}}|=\color{red}{\frac{\sqrt 6}{3}}$$