Let $ a_{1}, a_{2}...a_{n}$ be non-negative reals, not all zero. Show that that (a) The polynomial $ p(x) = x^{n} - a_{1}x^{n - 1} + ... - a_{n - 1}x - a_{n}$ has preceisely 1 positive real root $ R$. (b) let $ A = \sum_{i = 1}^n a_{i}$ and $ B = \sum_{i = 1}^n ia_{i}$. Show that $ A^{A} \leq R^{B}$.
Problem
Source: 1996 IMO Shortlist
Tags: inequalities, algebra, polynomial, function, calculus, IMO Shortlist
silouan
15.06.2006 16:21
a)Consider the function $f(x)=\sum_{i=1}^n\frac{a_i}{x^i}$ and this function is decreasing for positive x .So there is exactly one $R$ sucth that $f(R)=1$. b)Also by Jensen inequality we have $\sum_{i=1}^n\frac{a_i}{A}(ln\frac{A}{R^i})\leq ln(\sum_{i=1}^n\frac{a_i}{A}\frac{A}{R^i})=ln\ f(R)=0$ So $AlnA\leq BlnR$
mathisfun1
24.06.2006 17:42
Does a) follow directly from Descartes' rule of signs? Perhaps I am missing some detail...
CatalystOfNostalgia
07.06.2009 23:53
Yes, Descartes will work but admittedly I used Descartes on a problem like three days ago but didn't think to use it. Anyway, for (a) you can also induct on $ n$ and apply Rolle's Theorem, we have that the derivative has at most one positive root, so the polynomial has at most two, but it's not hard to show that you need an odd number of positive roots (counting multiplicity) since f(0)<0 and f(x) goes to positive infinity. And obviously, this implies that there is at least one root as well. For (b), we can show that $ f(A^{A/B})$ is negative by dividing both sides by $ A$ and applying weighted AM-GM. Then, for x>R. we have f(x)>0 or else we get another root bigger than R.