All zeroes of monic cubic $x^3+ax^2+bx+c$ are negative reals and $a\lt3$. Range of $b+c$?

Question

$a,b,c$ are real numbers. I have to find the range of values of $b+c$.

So, I started off by assuming $\alpha , \beta , \gamma$ as the roots. This gives us $\alpha \beta \gamma = -c$ and $\sum{\alpha\beta}=b$.

From the above two equations, we can conclude that both $b$ and $c$ are positive.

Also, $\alpha + \beta + \gamma \gt -3$ since the sum of roots is equal to $-a$. Now, if we consider $\alpha' , \beta' , \gamma'$ as the positive counterparts of our three roots, we can use the AM GM inequality. We get: $\frac{\alpha'+\beta'+\gamma'}3\ge(\alpha' \beta'\gamma')^{1/3}$ Replacing values in above equation, we get $\frac a3 \ge c^{1/3}$.

Since $a<3$, we get $c<1$.

That's where I'm stuck. I have no idea how to get the range of values for $b$. Any help is appreciated.

Answer 1

The derivative $3x^2+2ax+b$ has also two real, negative roots. Then $$4a^2-12b\ge0$$

I hope this helps.

Answer 2

Note that $(\alpha-1)(\beta-1)(\gamma-1)=-1-a-b-c$.

So, we need to find the range of $(\alpha-1)(\beta-1)(\gamma-1)$.

Let $\alpha', \beta',\gamma'$ be absolute values of the roots.

Hence, $|(\alpha-1)(\beta-1)(\gamma-1)|=(\alpha'+1)(\beta'+1)(\gamma'+1)$.

So, let's find the bounds of the $(\alpha'+1)(\beta'+1)(\gamma'+1)$.

Since roots are negative, we have $\alpha'>0, \beta'>0,\gamma'>0$, or

$(\alpha'+1)(\beta'+1)(\gamma'+1)>1$. That's lower bound.

Since a<3, we have $\alpha'+ \beta'+\gamma'<3$, so

$(\alpha'+1)(\beta'+1)(\gamma'+1)\leq(\frac{\alpha'+1+\beta'+1+\gamma'+1}{3})^3<8$.

Hence, we have that bounds are $-1$ and $-8$, so:

$$-1>(\alpha-1)(\beta-1)(\gamma-1)>-8$$ $$-1>-1-a-b-c>-8$$ $$-a<b+c<7-a$$ and since $b+c>0$, $a>0$ always, we have $$0<b+c<7-a$$