Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $-\frac{7}{3}<a+b<-2$. Proposed by Serbia
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Tags: algebra, inequalities, JBMO
Tintarn
12.09.2020 14:20
Let $s=a+b$ and $p=ab$. Then the equation becomes $s(s^2-3p)-6p=-11$. Hence $p=\frac{s^3+11}{3s+6}$ (in particular $s \ne -2$).
But $0 \le (a-b)^2=s^2-4p$ and hence
\[s^2 \ge \frac{4(s^3+11)}{3s+6}.\]So letting $f(x)=x^3-6x^2+44$ we immediately find that $s>-2$ and $f(s)<0$ or $s<-2$ and $f(s)>0$.
Since $f'(x)=3x^2-12x=3x(x-4)$ and $f(4)=12>0$ we find that $f(x)>0$ whenever $x>x_0$ where $x_0$ is the unique real zero of $f$ and satisfies $x_0<0$.
Moreover, since $f(-2)=12>0$ we have $x_0<0$ and hence $s>-2$ and $f(s)<0$ cannot happen.
Hence we find that $s \in (x_0,-2)$ (and this is actually the sharpest result possible, i.e. all values of $s=a+b$ in this interval are actually possible).
To finish we just need to note that $x_0>-\frac{7}{3}$ or equivalently check that $f\left(-\frac{7}{3}\right)<0$ which is easy.
Afo
30.09.2020 17:33
Let $a+b=p$ and $ab=q$. The question becomes $$p^3-3pq-6q=-11$$ And so, $$q=\frac{11+p^3}{3(p+2)}$$ By AM-GM $$p^2 \ge 4q = \frac{4(11+p^3)}{3(p+2)}$$ $$\iff \frac{p^3-6p^2+44}{p+2}\leq 0$$ This can be finished by usual calculus. @Below, Fixed.
Tintarn
30.09.2020 18:36
Afo wrote: $$p^2 \ge 2q = \frac{2(11+p^3)}{3(p+2)}$$ $$\iff 3p^3+6p^2-22\ge 0$$ Are you sure? I think you missed $2p^3$ coming from the RHS in the expansion so you only get \[p^3+6p^2-22.\]More importantly, you have to be careful when multiplying with the denominator since for $p<-2$ it is negative. In fact, the inequality $p^2 \ge \frac{2(11+p^3)}{3(p+2)}$ is true for all $p<-6$ so you do not possibly get the desired result.
sqing
14.10.2020 04:18
$$-3=a^3+b^3+2^3-3ab\cdot 2=\frac{1}{2}(a+b+2)((a-2)^2+(b-2)^2+(a-b)^2)$$$$a+b<-2$$$$(a-2)^2+(b-2)^2+(a-b)^2\geq(a-2)^2+(b-2)^2 \geq\frac{(a+b)^2}{2}-4(a+b)+8> 18$$$$a+b=-\frac{6}{(a-2)^2+(b-2)^2+(a-b)^2}-2>-\frac{7}{3}$$
laks9985
15.10.2020 16:00
How you got>18
Pluto04
30.10.2020 20:55
laks9985 wrote: How you got>18 Using $$a+b<-2$$.
OlympusHero
22.07.2021 19:52
We have $(a+b)((a+b)^2-3ab)-6ab=-11$. Let $a+b=x, ab=y$. Then we have $x(x^2-3y)-6y=-11$, or $x^3-3xy-6y=-11$. We get $y(3x+6)=x^3+11$, so $y=\frac{x^3+11}{3x+6}$. Now we note that the solutions to $a+b=x, ab=y$ are the roots to the quadratic $z^2-xz+y$, so we must have $x^2-4y \geq 0$. Plugging in gives $x^2-\frac{4x^3+44}{3x+6} \geq 0$, or $x^2 \geq \frac{4x^3+44}{3x+6}$. We now have two cases; either the denominator is positive or negative. In the first case, we have $x^2(3x+6) \geq 4x^3+44$, or $3x^3+6x^2 \geq 4x^3+44$, which means $x^3-6x^2+44 \leq 0$. Let's try to get a rough bound on the solutions. We see that $-2$ is too big but $-2.5$ works, so we try $-2.3$. It doesn't work but is extremely close, and clearly everything below that works. However all of these make $3x+6$ negative, which is a contradiction. If the denominator is negative, then $x^2(3x+6) \leq 4x^3+44$, so $3x^3+6x^2 \leq 4x^3+44$, or $x^3-6x^2+44 \geq 0$. Similar to the above, the solutions are about $[-2.3, \infty)$, but since $3x+6<0$ we have $x<-2$ and hence our set is about $-2.3 < x < -2$. This satisfies $-\frac{7}{3}<x<-2$, as desired.
FriIzi
18.05.2023 13:54
Afo wrote: Let $a+b=p$ and $ab=q$. The question becomes $$p^3-3pq-6q=-11$$ And so, $$q=\frac{11+p^3}{3(p+2)}$$ By AM-GM $$p^2 \ge 4q = \frac{4(11+p^3)}{3(p+2)}$$ $$\iff \frac{p^3-6p^2+44}{p+2}\leq 0$$ This can be finished by usual calculus. @Below, Fixed. maybe true
sqing
18.05.2023 15:20
Let $a,b$ be real numbers satisfy $a+b\leq -3.$ Prove that $$a^3+b^3-6ab\leq-\frac{81}{4}$$
FriIzi
18.05.2023 15:52
sqing wrote: Let $a,b$ be real numbers satisfy $a+b\leq -3.$ Prove that $$a^3+b^3-6ab\leq-\frac{81}{4}$$ $p=a+b$ $q=ab$ $p^3-3pq-6q \le \frac{-81}{4}$ $4p^3-12pq-24q \le -81$ $4p^3 \le -108$ $-108-12pq-24q \le -81$ $-12q(p+2) \le 27$ $-12q(p+2) \le -12q(-3+2)=12q \le 27 (?)$ $q \le \frac{9}{4}$ note that $4q \le p^2 \le 9$ and we are done
mudok
19.05.2023 14:11
I will continue Afo's solution. Afo wrote: Let $a+b=p$ and $ab=q$. The question becomes $p^3-3pq-6q=-11 \Rightarrow \frac{11+p^3}{3(p+2)}=q\le \frac{p^2}{4}$ $\iff \frac{p^3-6p^2+44}{p+2}\leq 0$ $0\ge \frac{p^3-6p^2+44}{p+2}= (4-p)^2+\frac{12}{p+2}\ge \frac{12}{p+2}$ $ \Rightarrow \frac{12}{p+2}\le 0 \Rightarrow p<-2 \Rightarrow \ 4-p>6 \Rightarrow\ (4-p)^2>36$ $0\ge (4-p)^2+\frac{12}{p+2}>36+\frac{12}{p+2}=12\cdot \frac{3p+7}{p+2}$ $\Rightarrow \frac{3p+7}{p+2}< 0 \ \ (\text{since} \ p+2<0) \Rightarrow 3p+7>0 \Rightarrow p>-\frac{7}{3}$