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: Showing that $a+b > - \frac73$:
$$(a-2)^2+(b-2)^2+(a-b)^2\geq(a-2)^2+(b-2)^2 \geq\frac{(a+b)^2}{2}-4(a+b)+8$$
Now using the previous result $a + b < -2$ we get:
$$\frac{(a+b)^2}{2}-4(a+b)+8> 18$$
From OP's equation in the question, this gives
$$a+b=-\frac{6}{(a-2)^2+(b-2)^2+(a-b)^2}-2>-\frac{6}{18} -2 = -\frac{7}{3}. \qquad \Box$$
A: 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.

A: $$a^3-6 a b+b^3=-11$$
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.
A: Let $f(a,b)=a^3+b^3-6ab+11$. At the extrema, the two curves $f(a,b)=0$ and $k=a+b$ are tangential to each other, i.e.
$$\frac{f_a’}{f_b‘}=\frac{3a^2- 6b}{3b^2-6a}=1\implies (a-b)(a+b+2)=0
$$
which leads to the tangential points $a=b$ and $a+b=-2$. It is straightforward to verify the maximum $a+b<-2$ and the minimum is determined by $f(a,a)=0$, or $2a^3-6a^2+11=0$, whose sole real root is given by the Cardano’s formula and is greater than $-\frac76$.
