Prove that $1 \leq A \leq \frac{5}{4}$ and $0 \leq B < \frac{81}{16}$ It’s known that A and B are real numbers it is also known that the polynomial P(x) has 4 real roots
$$P (x) = x^4 − 3x^3 + 3x^2 − Ax + B$$
I did come up with a solution for A and I was hoping to apply it to B but I just couldn’t.
If a polynomial has 4 roots it has 3 extremums. These extremums have different signs. That means that the derivative has 3 roots.
$$p’(x) = 4x^3-9x^2+6x-A$$
If a derivative has 3 roots it has 2 extremums (different signs). The second derivative has 2 roots.
$$p’’(x) = 12x^2-18x+6$$
The roots are 1 and $\frac{1}{2}$.
$$p’(1) = 1 -A$$
$$p’(1/2) = 5/4 - A$$
These extremums have to have different signs.
$$(1-A)(5/4-A)\leq 0$$
To apply the same for B I have to find the roots of the derivative.
I also came up with a solution for one part of B but really don’t like it (the limitation is stronger than required and it is kinda sloppy).
$$B = {x_1}{x_2}{x_3}{x_4} $$
$$3={x_1}+{x_2}+{x_3}+{x_4}$$
$$3 = {x_1}{x_2}+…+{x_3}{x_4}$$
$$9= {x_1}^2+{x_2}^2+{x_3}^2+{x_4}^2+2( {x_1}{x_2}+…+{x_3}{x_4})$$
$$3= {x_1}^2+{x_2}^2+{x_3}^2+{x_4}^2$$
Suppose ${y_i} = abs({x_i})$
$$3 = {y_1}^2+{y_2}^2+{y_3}^2+{y_4}^2 $$
$$sqrt[2] {\frac{{y_1}^2+{y_2}^2+{y_3}^2+{y_4}^2}{4}} \geq \sqrt[4] {{y_1}{y_2}{y_3}{y_4}}$$
$$ \frac{81}{16} \geq \frac{9}{16} \geq {y_1}{y_2}{y_3}{y_4} \geq {x_1}{x_2}{x_3}{x_4}$$
 A: Let $a,b,c,d$ be the roots of $P$.
First we show the bound of $A$.
By Rolle's theorem, we know that $P'(x)=4x^3-9x^2+6x-A$ has at least $3$ roots.
At the same time, $P''(x)=12x^2-18x+6=6(x-1)(2x-1)$
We can easily see that $P'(x)$ is increasing in $]- \infty,\frac1 2]$ decreasing in $[\frac1 2,1]$ and increasing in $[1, +\infty[$.
However, these variations imply that $P'$ has a root in $]- \infty,\frac1 2]$, we call it $\alpha$ (because $\lim_{x \to -\infty} P'(x)=-\infty$,$\lim_{x \to +\infty} P'(x)=+\infty$ and $P'$ must cross at least $3$ times the $x$-axis)
Then $P'(\frac1 2)\geqslant P'(\alpha)=0 \implies \frac 5 4-A \geqslant 0$, hence $A \leqslant \frac 5 4$ as $P'$ increasing here.
Secondly, we prove the other bound in a different way,using Vieta's formula as $B$ can easily be expressed this way.
We have $a+b+c+d=3$ and $ab+ac+ad+bc+bd+cd=3$ and $B=abcd$.
The bound we want to show is $\left(\frac 3 2 \right)^4$, so if we can show that all roots are positive roots less than $\frac 3 2$, this will conclude.
Thus we now see that $(a+b+c+d)^2-2(ab+ac+ad+bc+bd+cd)=a^2+b^2+c^2+d^2=3$
To deal with this sum, we use Cauchy-Schwarz inequality to bound the roots:
$3-a^2=b^2+c^2+d^2 \ge \frac{(b+c+d)^2}3=\frac{(3-a)^2}3$.
But, if $f(x)=3-x^2-\frac{(3-x)^2}3$ is a quadratic function with is positive if $x\in [0,\frac 3 2]$
This leads to the bound we wanted.
(I didn't see your edit while writing...)
A: Denote the roots by $r_1,r_2,r_3,r_4$, and let $e_i$ denote the $i^{\text{th}}$ elementary symmetric polynomial on the roots. By Vieta's formula, we have
$$e_1=3,\quad e_2=3,\quad e_3=A,\quad e_4=B.$$
We may then maximize and minimize $e_3$ subject to the above conditions on $e_1$ and $e_2$ using Lagrange multipliers, for example, to find
$$1\leq A\leq \frac{5}{4}\quad \textrm{and}\quad 0\leq B\leq \frac{3}{16}.$$
(Note my upper bound for $B$ is significantly tighter than the one you propose.) The minimum values are obtained when
$$P(x)=x(x-1)^3=x^4-3x^3+3x^2-x,$$
and the maximum values when
$$P(x)=\left(x-\frac{3}{2}\right)\left(x-\frac{1}{2}\right)^3=x^4-3x^3+3x^2-\frac{5}{4}x+\frac{3}{16}.$$
A: Another way to get estimations for $B$.
Let $a$, $b$, $c$ and $d$ are roots.
Thus, $$3=ab+ac+bc+ad+bd+cd\leq\frac{(a+b+c)^2}{3}+d(3-d)=\frac{(3-d)^2}{3}+d(3-d),$$
which gives $$2d^2-3d\leq0$$ or
$$0\leq d\leq\frac{3}{2},$$ which gives $$0\leq abcd\leq\frac{81}{16}$$ and $$0\leq B\leq\frac{81}{16}.$$
The right equality does not occur because if $a=b=c=d=\frac{3}{2}$, so $a+b+c+d=6,$ which is a contradiction,
which says $B<\frac{81}{16}.$
