$x^{3}+ax^2+bx+c$ has all roots negative real numbers and a<3. Establish an inequality between only b and c

Question

A cubic equation $x^{3}+ax^2+bx+c$ has all negative real roots and $a, b, c\in R$ with $a<3.$

Prove that $b+c<4.$

My attempt :

Let the cubic be $f(x)$

Plotting graph we see that ,

$f(x\geq 0)>0$.

So we can see that a relation between $a, b, c$ can be established by putting $x=1$, so,

$1+a+b+c> 0$.

So $b+c>-4$.

Also by using Vieta's Formula we get $a, b, c > 0$.

Now I'm uncertain how to proceed, any help will be appreciated. This is a problem of a maths olympiad. Thanks.

Answer 1

Let $-p,-r,-q$ be it roots, so $p,q,r>0$. Now by Vitea formulas we have $$p+r+q =a < 3$$ and $$b+c = pr+pq+qr +qpr$$

So we need to check if $$pr+pq+qr +qpr<4$$

By Am-Gm ineaquality we have $$pqr \leq ({p+q+r\over 3})^3 < 1$$

It is easy to check that (say with Cauchy inequality) $$p^2+q^2+r^2\geq pq+pr+qr$$ so $$(p+q+r)^2\geq 3(pq+pr+qr)$$ and thus $$pq+pr+qr \leq 3$$ so $$pq+pr+qr +pqr< 3+1=4$$

and we are done.

Answer 2

We can write our polynomial in the form $(x+r_1)(x+r_2)(x+r_3)$, where $r_i$ are positive real numbers. Note that $b=r_1r_2+r_2r_3+r_3r_1$ and $c=r_1r_2r_3$.

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$\color{Green}{\text{Maybe you will find}}$ Maclaurin's inequality $\color{Green}{\text{very interesting}}$.

By Maclaurin's inequality, we know that $\sqrt[3]{S_3} \leq \frac{S_1}{3}$, or equivalently: $$c=r_1r_2r_3 = S_3 \leq (\frac{S_1}{3})^3 = (\frac{a}{3})^3 \leq (\frac{3}{3})^3=1.$$

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Also again by use of Maclaurin's inequality, we know that $\sqrt[2]{\frac{S_2}{3}} \leq \frac{S_1}{3}$, or equivalently: $$b=r_1r_2+r_2r_3+r_3r_1 = S_2 \leq 3(\frac{S_1}{3})^2 = 3(\frac{a}{3})^2 \leq 3(\frac{3}{3})^3=3.$$

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so $$b+c \leq 1+3=4.$$

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$\color{Red}{\text{Comment}}$: Also, you can obtain all of the above inequalities $\color{Red}{\text{only}}$ by using AM–GM inequality.