IMO 1996 SL A4

Origin: KOR | Category: Algebra

Problem

Let $a_1, a_2, \dots, a_n$ be nonnegative real numbers, not all zero.

(a) Prove that

$$x^n - a_1 x^{n-1} - a_2 x^{n-2} - \cdots - a_{n-1} x - a_n = 0$$

has precisely one positive real root.

(b) Let

$$A = \sum_{j=1}^{n} a_j, \quad B = \sum_{j=1}^{n} j a_j,$$

and let $R$ be the positive real root of the equation in part (a). Prove that

$$A^A \leq R^B.$$

Solution

Consider the function

$$f(x) = \frac{a_1}{x} + \frac{a_2}{x^2} + \cdots + \frac{a_n}{x^n}.$$

Since $f$ is strictly decreasing from $+\infty$ to $0$ on the interval $(0, +\infty)$, there exists exactly one $R > 0$ for which $f(R) = 1$. This $R$ is also the only positive real root of the given polynomial.

Since $\ln x$ is a concave function on $(0, +\infty)$, Jensen’s inequality gives

$$\sum_{j=1}^{n} \frac{a_j}{A} \ln \frac{A}{R^j} \leq \ln \sum_{j=1}^{n} \frac{a_j}{A} \cdot \frac{A}{R^j} = \ln f(R) = 0.$$

Therefore

$$\sum_{j=1}^{n} a_j (\ln A - j \ln R) \leq 0,$$

which is equivalent to

$$A \ln A \leq B \ln R,$$

i.e.,

$$A^A \leq R^B.$$