Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $-\frac{7}{3}
Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $$-\frac{7}{3}<a+b<-2$$
I have shown that $a+b<-2$. My approach: $-3=8-11=a^3+b^3-6ab+2^3=\frac{1}{2}(a+b+2)((a-b)^2+(a-2)^2+(b-2)^2)$. From this we must have that $a+b<-2$.
Please give some idea/hint for the other part.

Hint:
Let $x=a+b$ then $b=x-a$ so we get an quadratic equation on $a$: $$3a^2(x+2)-3a(x^2+2x)+x^3+11=0$$ Since $a$ is real it discriminant is non negative, so we have $$-3(x+2)(x^3-6x^2+44)\geq 0$$

 Notice that $x\mapsto x^3$ and $x\mapsto -6x^2$ are increasing for $x \leq 0$.
 Now if $x\leq -{7\over 3}$ then we get $$\Big(-{7\over 3}\Big)^3-6 \Big(-{7\over 3}\Big)^2 +44\geq 0$$ A contradicition.

