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

Revision as of 05:30, 23 October 2023

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 $k=1$ . $\blacksquare$

Solution 1

$a_1+a_2+.....+a_k \leq \sqrt{n}.\sqrt{a_1^2+a_2^2+................+a_k^2}$ For all real numbers. $a_1,a_2,....$ Hence it is only required to prove $a_1^2+a_2^2+................+a_k^2<1$ where $a_k=\dfrac{x_k}{1+x_1^2+... x_k^2}$

for $k \geq 2$ , $a_k^2=(\dfrac{x_k}{1+x_1^2+... x_k^2})^2 \leq \dfrac{x_k^2}{(1+x_1^2+... x_{k-1}^2)(1+x_1^2+... x_k^2)} = \dfrac{1}{1+x_1^2+... x_{k-1}^2}-\dfrac{1}{1+x_1^2+... x_k^2}$

For k=1 $a_1^2\leq1-\dfrac{1}{1+x_1^2}$

Summing these inequalities, the right-hand side yields $\sum_{n=1}^{k}a_n^2\leq 1$

Hence Proved by Maths1234RC. $\blacksquare$

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

Resources

@@ Line 25: / Line 25: @@
 <cmath> (\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, </cmath>
 with equality only when <math>(\lvert \sin t \rvert, \cos t) = (1/\sqrt{n^2+1}, n/\sqrt{n^2+1}</math>.  Since <math>\left\lvert n/\sqrt{n^2+1} \right\rvert < 1</math>, our equality cases never coincide, so we have the desired strict inequality for <math>n+1</math>.  Thus our inequality is true by induction.  The problem statement therefore follows from setting <math>k=1</math>.  <math>\blacksquare</math>
+== Solution 1==
+<math>a_1+a_2+.....+a_k \leq \sqrt{n}.\sqrt{a_1^2+a_2^2+................+a_k^2}</math>
+For all real numbers. <math>a_1,a_2,....</math>
+Hence it is only required to prove <math>a_1^2+a_2^2+................+a_k^2<1</math> where <math>a_k=\dfrac{x_k}{1+x_1^2+... x_k^2}</math>
+for <math>k \geq 2</math>, <math>a_k^2=(\dfrac{x_k}{1+x_1^2+... x_k^2})^2 \leq \dfrac{x_k^2}{(1+x_1^2+... x_{k-1}^2)(1+x_1^2+... x_k^2)} = \dfrac{1}{1+x_1^2+... x_{k-1}^2}-\dfrac{1}{1+x_1^2+... x_k^2}</math>
+For k=1 <math>a_1^2\leq1-\dfrac{1}{1+x_1^2}</math>
+Summing these inequalities, the right-hand side yields
+<math>\sum_{n=1}^{k}a_n^2\leq 1</math>
+Hence Proved by Maths1234RC.<math>\blacksquare</math>

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