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

Latest revision as of 23:54, 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 1

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 2

By the Cauchy-Schwarz Inequality $a_1+a_2+.....+a_k \leq \sqrt{n}.\sqrt{a_1^2+a_2^2+................+a_k^2}$ 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 $k \geq 2$ , $a_k^2=(\dfrac{x_k}{1+x_1^2+... x_k^2})^2 < \dfrac{x_k^2}{(1+x_1^2+... x_{k-1}^2)(1+x_1^2+... x_k^2)} = \dfrac{1}{1+x_1^2+... x_{k-1}^2}-\dfrac{1}{1+x_1^2+... x_k^2}$

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

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

Hence Proved by Maths1234RC P.S. This is my first solution on AOPS. $\blacksquare$

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

Resources

@@ Line 32: / Line 32: @@
 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 <math>k \geq 2</math>, <math>a_k^2=(\dfrac{x_k}{1+x_1^2+... x_k^2})^2 < \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>
+For k=1, <math>a_1^2 = 1-\dfrac{1}{1+x_1^2}</math>
 Summing these inequalities, the right-hand side yields

Page

Toolbox

Search