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.
Show that \(pqr,qrs,rsp\) and \(spq\) are distinct, where \(p,q,r\) and \(s\) are the roots of the polynomial \(\mathrm{f}\).
By considering the relationship between the coefficients of \(\mathrm{f}\) and its roots, show that \(c > d\).
Explain why the polynomial \(\mathrm{f}'(x)\) must have three distinct roots.
By differentiating \(\mathrm{f}\), show that \(b > c\).
Show that \(a > b\).
Solution:
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.
\(4c^3 = pqr+qrs+rsp+spq\), \(d^4 = pqrs\).
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*}
There must be a turning point between each root (since there are no repeated roots).
\(f'(x) = 4x^3-12ax^2+12b^2-4c^3 = 4(x^3-3ax^2+3b^2-c^3)\). 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*}
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 \(a>b\)