What does mean SMO??

= Singapore Math Olympiad

roza2010 wrote: = Singapore Math Olympiad I don't think so. See here: http://www.artofproblemsolving.com/Forum/resources.php?c=150&cid=237&year=2014&sid=eb5a0670ada206e50c47f5d41dec5a97

singapore math olympiad open. It's a 3-hour-5-question olympiad

I have the impression that this must be a strict inequality, the equality is never achieved. Am I right?

Let $b_k = 1+a_k$ ($k=1..n$). We have to prove that \[\left (\frac{1}{b_1}+\frac{1}{b_2}+\cdots +\frac{1}{b_n}\right )^2 \leq \frac{1}{b_1 - 1}+\frac{1}{b_2-b_1}+\cdots +\frac{1}{b_n-b_{n-1}} \quad (*)\]We prove by induction that (*) is true for all $1 < b_1 < \ldots < b_n$ ($n \geq 1$). For $n = 1$ (*) is obvious, suppose now (*) is true for all $1 < b_1 < \ldots < b_{n-1}$ ($n \geq 2$). Let $x = b_1 > 1$ and $q_k= \frac{b_k}{x}$ ($k=2..n$) so $1<q_2< \ldots < q_n$. We can write (*) as $$\frac{1}{x^2} A \leq \frac{1}{x-1} + \frac{1}{x}B \ \text{ where } \ A = \left( 1 + \frac{1}{q_2} + \ldots +\frac{1}{q_n}\right)^2 \ \text{, } B = \left( \frac{1}{q_2 - 1} + \frac{1}{q_3 - q_2}+ \ldots +\frac{1}{q_n - q_{n-1}}\right)$$that is, $(1+B)x^2 -x(A+B) + A \geq 0 \Leftrightarrow x^2 +(x-1)(Bx-A) \geq 0$. If $B \geq A$ then $Bx-A \geq B-A \geq 0$ and the inequality is obvious. Assume in the following that $B < A$. It's enough to show that the discriminant $\Delta = (A+B)^2 - 4A(1+B) = (A-B)^2 - 4A$ is negative, that is $(A-B)^2 - 4A < 0 \Leftrightarrow A-B < 2 \sqrt A$ (because $A>B$). But we can apply the induction hypothesis to the $(n-1)$ tuple $(q_2, \ldots q_n)$ hence $B \geq \left(\frac{1}{q_2} + \ldots +\frac{1}{q_n}\right)^2$. Therefore $ A = \left( 1 + \frac{1}{q_2} + \ldots +\frac{1}{q_n}\right)^2 = 1 + 2\left(\frac{1}{q_2} + \ldots +\frac{1}{q_n}\right) + \left(\frac{1}{q_2} + \ldots +\frac{1}{q_n}\right)^2$ $\leq 1 + 2\left(\frac{1}{q_2} + \ldots +\frac{1}{q_n}\right) + B < 2\left(1 + \frac{1}{q_2} + \ldots +\frac{1}{q_n}\right) + B = 2\sqrt A + B$ hence $A - B < 2 \sqrt A$ etc.

\bump any one-liners?

Konigsberg wrote: Let $0<a_1<a_2<\cdots <a_n$ be real numbers. Prove that \[\left (\frac{1}{1+a_1}+\frac{1}{1+a_2}+\cdots +\frac{1}{1+a_n}\right )^2 \leq \frac{1}{a_1}+\frac{1}{a_2-a_1}+\cdots +\frac{1}{a_n-a_{n-1}}.\] By C-S, $LHS\le RHS\cdot \sum \frac{a_i-a_{i-1}}{(1+a_i)^2}$, and note that $\sum \frac{a_i-a_{i-1}}{(1+a_i)^2}<\sum \frac{a_i-a_{i-1}}{(1+a_i)(1+a_{i-1})}=\sum \left(\frac{1}{1+a_{i-1}}-\frac{1}{1+a_i}\right)=1-\frac{1}{1+a_n}<1$, which gives that $LHS\le RHS$. yellowhanoi wrote: I have the impression that this must be a strict inequality, the equality is never achieved. Am I right? Actually it turns out to be quite loose