2001 IMO Shortlist Problems/A3

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. $\blacksquare$

Solution 3

Let $S_n = 1 + x^2 + \cdots + x_n^2$ where $S_0 = 1$ by convention.

Rewriting the inequality, $\frac{x_1}{S_1} + \frac{x_2}{S_2} + \cdots + \frac{x_n}{S_n} < \sqrt{n}$ must be proven.

Note that by Chebyshev's inequality, the following could be driven. $\left( \sum_{i = 1}^n \frac{x_i}{S_i} \right)^2 \leq n \left( \sum_{i = 1}^n \frac{x_i^2}{S_i^2} \right)$ Therefore, it suffices to prove that $\sum_{i = 1}^n \frac{x_i^2}{S_i^2} < 1$ . Notice that $x_i^2 = S_i - S_{i - 1}$ .

 
 
 is true.

$Proof.$ Because $S_n$ is a monotonically increasing sequence, $S_i^2 \geq S_i \cdot S_{i - 1}$ . $\sum_{i = 1}^n \frac{S_i - S_{i - 1}}{S_i^2} \leq \sum_{i = 1}^n \frac{S_i - S_{i - 1}}{S_i \cdot S_{i - 1}} = \sum_{i = 1}^n \left( \frac{1}{S_{i - 1}} - \frac{1}{S_i} \right) = \frac{1}{S_0} - \frac{1}{S_n} = 1 - \frac{1}{S_n} < 1$ $\blacksquare$

Hence, the inequality is true.

~MaPhyCom (Thinking Tree)

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