<s>[INST] 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.
 [/INST] 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 $-\frac76$.
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