Let $ a_1,a_2,\dots$ be sequence of real numbers such that $ a_1=1$, $ a_2=\dfrac{4}{3}$, and \[ a_{n+1}=\sqrt{1+a_na_{n-1}}, \quad \forall n \ge 2.\] Prove that for all $ n \ge 2$, \[ a_n^2>a_{n-1}^2+\dfrac{1}{2}\] and \[ 1+\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dots+\dfrac{1}{a_n}>2a_n.\] Fajar Yuliawan, Bandung
Problem
Source: Indonesia IMO 2010 TST, Stage 1, Test 3, Problem 3
Tags: inequalities, induction, algebra proposed, algebra
Raja Oktovin
14.11.2009 06:42
The first inequality.
We will prove by induction.
The initial cases are trivial and assume that we may use them now. Note that $ a_n$ are positive reals.
If this inequality is true for $ k$ and $ k-1$, then $ a_k^2>a_{k-1}^2+\dfrac{1}{2}$ and $ a_{k-1}^2>a_{k-2}^2+\dfrac{1}{2}$, then by Cauchy Schwarz Inequality \[ a_{k+1}^2=1+a_ka_{k-1}>1+\sqrt{\left(a_{k-1}^2+\dfrac{1}{2}\right)\left(a_{k-2}^2+\dfrac{1}{2}\right)} \ge 1+a_{k-1}a_{k-2}+\dfrac{1}{2}=a_k^2+\dfrac{1}{2}.\]
Q.E.D.
The second inequality.
Note that from the first inequality, we obtain \[ \dfrac{1}{a_n}>2(a_n-a_{n-1})\] but this is telescoping series and the next is straightforward. Q.E.D.
Any solution?
Moubinool
14.11.2009 20:24
Could you post the detail for the second inequality thank's
leshik
14.11.2009 21:19
Well, second inequality immediately follows from the first one (just observe that $ \frac{1}{a_n}> 2(a_n-a_{n-1})$ is eqvivalet to $ 1+2a_n a_{n-1}> 2a_n ^2$ or,using reccurence formula, $ 2a_{n+1} ^2 -1>2 a_n ^2$ which is exactly the first inequality from the problem).
Moubinool
14.11.2009 21:48
What is an equivalent of $ a_{n}$ ?
leshik
14.11.2009 22:06
It is obvious that the order of our sequence is $ \sqrt{n}$ (the lower bound follows from simple induction and the first inequality,the upper bound comes from simple induction).
nawaites
14.07.2014 10:51
Any solution to these. Problem?