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.

$$a^3-6 a b+b^3=-11$$
can be written as
$$(a+b)^3-3 a b (a+b)-6 a b=-11$$
setting $$a+b=s;\;ab=p$$
we get
$$ s^3-3ps-6p+11=0$$
The equation $$z^2-sz+p=0\tag{1}$$
Gives the values of $a,b$. In order to have  real roots, discriminant must be positive
$$s^2-4p\ge 0$$
Solve
$$\begin{cases}
s^3-3ps-6p+11=0\\
s^2-4p\ge 0\\
\end{cases}
$$
Solving the first equation wrt $p$ we get
$$p=\frac{s^3+11}{3 (s+2)}$$
From the second equation we get $p\le \frac{s^2}{4}$.
Substituting in the first we get the inequality
$$\frac{s^3+11}{3 (s+2)}\le  \frac{s^2}{4}$$
$$\frac{s^3+11}{3 (s+2)}=  \frac{s^2}{4}\to s^3-6 s^2+44=0\to s=-2.30214$$
The solution of the inequality is
$$-2.30214 \le s<-2$$
that is
$$-2.30214 \le a+b<-2$$
edit
which is a more accurate result than
$$-\frac73< a+b <-2$$
indeed if $s=-2.32>-\frac73$ then $p=1.54913$ and the equation $(1)$ becomes
$$z^2+2.32 z+1.54913=0$$
In other words the claim is wrong because values of $a+b$ from $-\frac73$ to $-2.30214$ the numbers $a$ and $b$ can't be real.
which has no real solutions because discriminant is negative.
