Problem source

In this question, you may assume that if $k_1,\dots,k_n$ are distinct positive real numbers, then
\[\frac1n\sum_{r=1}^nk_r>\left({\prod\limits_{r=1}^n}
k_r\right )^{\!\! \frac1n},\]
i.e. their arithmetic mean is greater than their geometric mean.
Suppose that $a$, $b$, $c$ and $d$ are positive real numbers such that the polynomial
\[{\rm f}(x)=x^4-4ax^3+6b^2x^2-4c^3x+d^4\]
has four distinct positive roots.
\begin{questionparts}
\item Show that $pqr,qrs,rsp$ and $spq$ are distinct, where $p,q,r$ and $s$ are the roots of the polynomial $\mathrm{f}$.
\item By considering the relationship between the coefficients of $\mathrm{f}$ and its roots,  show that $c > d$.
\item Explain why the polynomial $\mathrm{f}'(x)$ must have three distinct roots. 
\item By differentiating $\mathrm{f}$, show that $b  > c$.
\item  Show that $a > b$.
\end{questionparts}

Solution source

\begin{questionparts}
\item Suppose $pqr = qrs$, since the roots are positive, we can divide by $qr$ to obtain $p=s$ (a contradiction. Therefore all those terms are distinct.

\item $4c^3 = pqr+qrs+rsp+spq$, $d^4 = pqrs$.

Applying AM-GM, we obtain:

\begin{align*}
&& c^3 = \frac{ pqr+qrs+rsp+spq}{4} & > \sqrt[4]{p^3q^3r^3s^3} = d^{3} \\
\Rightarrow && c &> d
\end{align*}

\item There must be a turning point between each root (since there are no repeated roots).

\item $f'(x) = 4x^3-12ax^2+12b^2-4c^3 = 4(x^3-3ax^2+3b^2-c^3)$. Letting the roots of this polynomial be $\alpha, \beta, \gamma$ and again applying AM-GM, we must have:

\begin{align*}
&& b^2 = \frac{\alpha\beta + \beta \gamma+\gamma \alpha}{3} &> \sqrt[3]{\alpha^2\beta^2\gamma^2} = c^2 \\
\Rightarrow && b &> c
\end{align*}

\item Again, since there are turning points between the roots of $f'(x)$ we must have distinct roots for $f''(x)$, ie:

$f''(x) = 3x^2-6ax+6b^2 = 3(x^2-2ax+b^2)$ has distinct real roots. But for this to occur we must have that $(2a)^2-4b^2 = 4(a^2-b^2) > 0$, ie $a>b$

\end{questionparts}