Cubic with negative real roots — inequality on coefficients

Question

Let $a, b, c \in \mathbb R$ such that all roots of $x^3 + a x^2 + b x + c$ are negative real numbers. If $a < 3$, prove that $b + c < 4$.

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My attempt:

I tried first putting a as positive and seeing what happens, which didn't really lead to anything. If we put $a$ as negative, then we have 3 terms which are certainly negative, which would imply that $c$ is certainly positive. this would also imply, that $|c| > |b|$

Now, we could say, since$ b $would be negative, that $b + c = c - |b|$. I have no idea now, how to prove that the difference between the two should be less than 4, and also how to prove that this should be true when $a$ is positive but less than 3, since the problem hasn't provided a clue on the possible roots of the equation.

Answer 1

Let $-u$, $-v$ and $-w$ be our roots.

Thus, $u$, $v$ and $w$ are positives, by the given $u+v+w<3$ and we need to prove that: $$uv+uw+vw+uvw<4.$$ Now, let $u=tp$, $v=tq$ and $w=tr$ such that $t>0$ and $p+q+r=3.$

Thus, $$t(p+q+r)<3,$$ which gives $$0<t<1.$$ But $$uv+uw+vw+uvw=t^2(pq+pr+qr)+t^3pqr<pq+pr+qr+pqr$$ and it's enough to prove that: $$pq+pr+qr+pqr\leq4$$ or $$\frac{(pq+pr+qr)(p+q+r)}{3}+pqr\leq\frac{4(p+q+r)^3}{27}$$ or $$\sum_{cyc}(4p^3+3p^2q+3p^2r-10pqr)\geq0,$$ which is true by AM-GM or by Muirhead or by very many another ways.

Answer 2

Similar to Michael's start, we WTS if $ u + v + w < 3$, then

$$uv+uw+vw+uvw<4.$$

This follows from:

$\sqrt[3]{uvw} \leq \frac{ u+v+w}{3} < 1 \Rightarrow uvw < 1$, and
$3 (uv + vw + wu ) \leq (u+v+w)^2 < 9 \Rightarrow uv+vw+wu < 3$.