If all the roots of $P(x)=x^3-\sqrt{27}x^2+bx-\sqrt{27}$ are real and positive, find the value of $b$ for $b>0$ If all the roots of $P(x)=x^3-\sqrt{27}x^2+bx-\sqrt{27}$ are real and positive, I'm trying to find the positive real value of $b$.
If the three roots are $r_1, r_2 $ and $r_3$ then $r_1+r_2+r_3=\sqrt{27}$, $r_1r_2r_3=\sqrt{27}$ and $r_1r_2+r_2r_3+r_1r_3=b$ using Vieta's formulas.
Using a well known inequality $r_1^2+r_2^2+r_3^2≥r_1r_2+r_2r_3+r_1r_3$ which can be re-written as $(r_1+r_2+r_3)^2≥3(r_1r_2+r_2r_3+r_1r_3)$. This gives me $b≤9$.
I wasn't really sure what to do from here so I just guessed. Looking at how symmetrical the equations $r_1+r_2+r_3=\sqrt{27}$ and $r_1r_2r_3=\sqrt{27}$ are, I guessed that $r_1=r_2=r_3=\sqrt{3}$ satisfies the set of equations and hence I get $b=9$, which fulfills the original conditions as it means $P(x)$ has a triple root at $x=\sqrt{3}$.
However, my question is can I prove that this is the only solution (if it is even the only solution? I have looked at graphs on desmos as I vary the value of $b$ and it seem s to me to be the only solution but I can't prove it. Could someone help me do this?
 A: Consider the function $$f(x)=-x^2+\sqrt{27}x+\frac{\sqrt{27}}x.$$ Roots of $P$ are solutions of $f(x)=b$. The number of roots of this equation changes at the local maxima and minima of $f$. Globally, $f$ is falling on $x>0$, so that for large $|b|$ there will be only one positive root. At the local extrema this changes by $2$, so that between the extrema there will be $3$ real positive roots.
The local extrema of $f$ are at the roots of the derivative. The derivative factors as
\begin{align}
f'(x)&=-2x+\sqrt{27}-\frac{\sqrt{27}}{x^2},\\
x^2f'(x)&=-2x^2(x-\sqrt3)+\sqrt3\left(x^2-3\right)
\\&=(x-\sqrt3)\left[-2x^2+\sqrt3x+3\right]
\\&=(x-\sqrt3)^2\left[-2x-\sqrt3\right].
\end{align}
So $f'$ has a double root at $x=\sqrt3$. This means that no local extrema exist on the positive half axis, $f$ is strictly monotonously falling, there will always only be one positive real root. In the case $b=f(\sqrt3)=9$, it is a triple root at $x=\sqrt3$, this is the only value of $b$ to satisfy the requirements. One could say that this is a limit case, where the local minimum and maximum fall together, reducing the interval of admissible $b$ to one single point.
A: Based on the "Vieta" system, we get:
$$r_1+r_2+r_3 = r_1r_2r_3$$
$$\implies r_1 = -\frac{(r_2+r_3)}{(1-r_2r_3)},\text{ }\text{ }\text{ if }r_2r_3 \neq 1.$$
$$\implies r_1 = -\frac{(r_2+r_3)}{(1-r_2r_3)},\text{ }\text{ }\text{ if }r_2 \neq \frac{1}{r_3}.$$
So any pair $(r_2,r_3)$ that are not multiplicative inverses, will determine $r_1$ by the RHS and thus solve the equation.

For such a triplet, we have the other equation via "Vieta" giving:
$$b = r_1r_2 + r_1r_3 + r_2r_3$$
$$ = -\frac{(r_2+r_3)}{(1-r_2r_3)}r_2 -\frac{(r_2+r_3)}{(1-r_2r_3)}r_3 + r_2r_3.$$$$ = -\frac{(r_2+r_3)^2}{(1-r_2r_3)}+r_2r_3.$$
As well, $b>0$ gives us:
$$r_2r_3 > \frac{(r_2+r_3)^2}{1-r_2r_3}.$$
This last condition can be translated to the graphical setting, where we compare surfaces:
$$f_1(x,y) = xy\text{ }\text{ }\text{ and }\text{ }\text{ }f_2(x,y) = \frac{(x+y)^2}{1-xy}.$$


In the desmos/paint image above, we see graphs giving insight to our problem, where $-f_1 < -f_2$ and such that $f_1 \neq 1$. (i.e. to where the black surface sits below the purple one and not on the intersection with the red one (should be shifted down to -1). The first octant is closest to us on the upper-left corner.
Therefore an educated guess would be say:
$$r_2 = 10\text{ }\text{ }\text{ and }\text{ }\text{ }r_3 = 11,\text{ }\text{ }\text{ which implies }\text{ }\text{ }r_1 = \frac{21}{109}.$$
We have:
$$r_1 + r_2 + r_3 = \frac{21}{109}+10+11 = \frac{21+1090+1199}{109} = \frac{2310}{109}$$
and
$$r_1+r_2+r_3 = \frac{21}{109}*10*11 = \frac{21*110}{109} = \frac{2310}{109}.$$
So we see this triplet satisfies $$r_1+r_2+r_3 = r_1r_2r_3.$$
Finally we have:
$$b = r_1r_2+r_1r_3+r_2r_3 = \frac{21}{109}10+\frac{21}{109}11+110$$
$$= \frac{210+231+110*109}{109} = \frac{12431}{109}>0.$$

$\therefore$ Since you've provided another triplet, there exists more than one $b>0$ for which $P(x)$ has 3 positive real solutions. This is of course provided this "Vieta" theorem is true, which you've used, but I haven't looked into.
A: Let' s multiply equation $f(x)=0$ with $3\sqrt{3}$ and let $t=\sqrt{3}x$. Then you get $$t^3-9t^2+3bt-27=0$$
So we want all $b$ such that equation $$ {t^3-9t^2-27 \over -3t}=b$$ has exactly 3 solutions. If we draw a graph for rational function on LHS we see that we have to calculate only local extremum and we get $t=-{3\over 2}$ (for max) and $t=3$ (for min) so for all $b< -{45\over 4}$ or $b= 3$ 
