1998 IMO Shortlist Problems/A2

Revision as of 21:03, 30 December 2007 by Boy Soprano II (talk | contribs) (New page: == Problem == (''Australia'') Let <math>r_1, r_2, \dotsc, r_n</math> be real numbers greater than or equal to 1. Prove that <cmath> \frac{1}{r_1 + 1} + \frac{1}{r_2 + 1} + \dotsb + \frac...)
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Problem

(Australia) Let $r_1, r_2, \dotsc, r_n$ be real numbers greater than or equal to 1. Prove that \[\frac{1}{r_1 + 1} + \frac{1}{r_2 + 1} + \dotsb + \frac{1}{r_n + 1} \ge \frac{n}{\sqrt[n]{r_1 r_2 \dotsm r_n} + 1} .\]

Solution

Let $f$ denote the function $x \mapsto 1/(e^x + 1)$.

Lemma 1. For $x \ge 1$, the function $g : x \mapsto x/(x +1)^2$ is decreasing.

Proof. Note that $g(x) = \frac{1}{x + 2 + 1/x}$. Since $x + 1/x$ is increasing for $x\ge 1$, the lemma follows. $\blacksquare$

Lemma 2. For positive $x$, $f(x)$ is convex.

Proof. Note that the derivative of $f$ is \[\frac{df}{dx} = - \frac{e^{x}}{(e^x+1)^2} .\] By Lemma 1, $df/dx$ is increasing when $e^x \ge 1$, i.e., when $x\ge 0$. Therefore $f$ is convex for nonnegative $x$. $\blacksquare$

For all integers $1 \le k \le n$, $r_k \ge 1$, so $\ln r_k \ge 0$. Since $f(x)$ is convex for nonnegative $x$, it follows from Jensen's Inequality that \[\sum_{k=1}^n \frac{1}{r_k+1} = \sum_{k=1}^n f(\ln r_k) \ge n \cdot f \left( \sum_{k=1}^n \ln r_k/n \right) = \frac{1}{ \prod_{k=1}^n r_k^{1/n} +1},\] as desired. $\blacksquare$


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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