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?
Best Answer
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.