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Difference between revisions of "2001 IMO Shortlist Problems/A3"

(New page: == Problem == Let <math>x_1, x_2, \dotsc, x_n</math> be arbitrary real numbers. Prove the inequality <cmath> \frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \dotsb + \frac{x_n}{1+ x_...)
 
m (typo)
Line 6: Line 6:
 
== Solution ==
 
== Solution ==
  
−
We prove the following general inquality, for arbitrary positive real <math>k</math>:
+
We prove the following general inequality, for arbitrary positive real <math>k</math>:
 
<cmath> \sum_{j=1}^n \frac{x_j}{k^2 + \sum_{i=1}^j x_i^2} \le \sqrt{n}/k , </cmath>
 
<cmath> \sum_{j=1}^n \frac{x_j}{k^2 + \sum_{i=1}^j x_i^2} \le \sqrt{n}/k , </cmath>
 
with equality only when <math>n=0</math>.
 
with equality only when <math>n=0</math>.

Revision as of 10:03, 29 December 2007

Problem

Let $x_1, x_2, \dotsc, x_n$ be arbitrary real numbers. Prove the inequality \[\frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \dotsb + \frac{x_n}{1+ x_1^2 + \dotsb + x_n^2} < \sqrt{n} .\]

Solution

We prove the following general inequality, for arbitrary positive real $k$: \[\sum_{j=1}^n \frac{x_j}{k^2 + \sum_{i=1}^j x_i^2} \le \sqrt{n}/k ,\] with equality only when $n=0$.

We proceed by induction on $n$. For $n=0$, we have trivial equality. Now, suppose our inequality holds for $n$. Then by inductive hypothesis, \[\sum_{j=1}^{n+1} \frac{x_j}{k^2 + \sum_{i=1}^j x_i^2} = \frac{x_1}{k^2 + x_1^2} + \sum_{j=2}^{n+1} \frac{x_j}{k^2 + x_1^2 + \sum_{i=2}^j x_i^2} \le \frac{x_1}{k^2 + x_1^2} + \frac{\sqrt{n}}{\sqrt{k^2 + x_1^2}} .\] If we let $t= \text{Arcsin} \left(x_1/\sqrt{x_1^2 +k^2} \right)$, then we have \[\frac{x_1}{k^2 + x_1^2} + \frac{\sqrt{n}}{\sqrt{k^2+x_1^2}} = (\sin t \cos t + \sqrt{n} \cos t)/k \le (\lvert \sin t \rvert + \sqrt{n} \cos t)/k ,\] with equality only if $\cos t= \pm 1$. By the Cauchy-Schwarz Inequality, \[(\lvert \sin t \rvert + \sqrt{n} \cos t)/k \le (1 + n)^{1/2}(\sin^2 t + \cos^2 t)^{1/2}/k = \sqrt{n+1}/k,\] with equality only when $(\lvert \sin t \rvert, \cos t) = (1/\sqrt{n^2+1}, n/\sqrt{n^2+1}$. Since $\left\lvert n/\sqrt{n^2+1} \right\rvert < 1$, our equality cases never coincide, so we have the desired strict inequality for $n+1$. Thus our inequality is true by induction. The problem statement therefore follows from setting $k=1$. $\blacksquare$


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