Consider two infinite sequences $a_0,a_1,a_2,\dots$ and $b_0,b_1,b_2,\dots$ of real numbers such that $a_0=0$, $b_0=0$ and \[a_{k+1}=b_k, \quad b_{k+1}=\frac{a_kb_k+a_k+1}{b_k+1}\]for each integer $k \ge 0$. Prove that $a_{2024}+b_{2024} \ge 88$.
Problem
Source: MEMO 2024 T1
Tags: Sequences, algebra, algebra proposed, inequalities, inequalities proposed
VedViesKto
28.08.2024 00:01
The key is the following:
Consider the value $(a_{k+1} + 1)(b_{k+1} + 1)$.
Solution:
It is easy to show that $(a_{k+1} + 1)(b_{k+1} + 1) = (a_k + 1)(b_k + 1) + 1$. Hence if we define the sequences $A_k = a_k + 1$ and $B_k = b_k + 1$, then $A_{k+1} B_{k+1} = A_k B_k + 1$. Since $A_0 = a_0 + 1 = 1$ and $B_0 = b_0 + 1 = 1$, we have $A_0 B_0 = 1 \cdot 1 = 1$. The forementioned gives $A_{2024} B_{2024} = 1 + 2024 = 2025$. Therefore using AM-GM we have
$$(a_{2024} + 1) + (b_{2024} + 1) = A_{2024} + B_{2024} \geq 2 \sqrt{A_{2024} \cdot B_{2024}} = 2 \sqrt{2025} = 90.$$Hence $a_{2024} + b_{2024} \geq 88$ as desired.
MathLuis
28.08.2024 01:18
First let's get rid of $b$ and modify our sequence condition as: $(a_{k+2}+1)(a_{k+1}+1)=(a_{k+1}+1)(a_k+1)+1$ for all $k \ge 0$. Since we are given that $a_0=a_1=0$ we get that $(a_2+1)(a_1+1)=2$ so inductively we can get that $(a_{k+1}+1)(a_k+1)=k+1$. Now by AM-GM $(a_{2025}+1)+(a_{2024}+1) \ge 2 \cdot \sqrt{(a_{2025}+1)(a_{2024}+1)}=2 \cdot \sqrt{2025}=90$ Which concludes, thus we are done .