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 ${y_i} = abs({x_i})$
$$3 = {y_1}^2+{y_2}^2+{y_3}^2+{y_4}^2 $$
$$sqrt[2] {\frac{{y_1}^2+{y_2}^2+{y_3}^2+{y_4}^2}{4}} \geq \sqrt[4] {{y_1}{y_2}{y_3}{y_4}}$$
$$ \frac{81}{16} \geq \frac{9}{16} \geq {y_1}{y_2}{y_3}{y_4} \geq {x_1}{x_2}{x_3}{x_4}$$
Best Answer
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 $B<\frac{81}{16}.$