Source: China Sijiazhuang , Aug 2014
Tags: inequalities proposed, inequalities, algebra, China
Toggle Solutions
Define a positive number sequence sequence $\{a_n\}$ by \[a_{1}=1,(n^2+1)a^2_{n-1}=(n-1)^2a^2_{n}.\]Prove that\[\frac{1}{a^2_1}+\frac{1}{a^2_2}+\cdots +\frac{1}{a^2_n}\le 1+\sqrt{1-\frac{1}{a^2_n}}
.\]
Click to reveal hidden text
Set $b_n = \frac{1}{a_n^2}$. Then we have $b_1=1$, $b_n(n^2+1) = b_{n-1}(n-1)^2$. We wish to prove that $\sum b_1 \leq 1 + \sqrt{1-b_n}$.
Note that $b_n = b_{n-1}(n-1)^2 - b_n(n^2)$ so the sum $b_1 + b_2 + \cdots + b_n$ telescopes:
\begin{eqnarray*} \sum_{i=1}^{n} b_i &=& b_1 + b_1 \cdot 1^2 - b_2 \cdot 2^2 + b_2 \cdot 2^2 - b_3 \cdot 3^2 \cdots + b_{n-1}(n-1)^2 - b_n(n^2) \\ &=& 2 - b_n \cdot n^2\end{eqnarray*}
It then suffices to show that $1-b_n \cdot n^2 \leq \sqrt{1-b_n}$. Note that $0 < b_n \leq 1 \Longrightarrow 0 \leq 1-b_n < 1$. We have $1-b_n \cdot n^2 \leq 1 - b_n \leq \sqrt{1-b_n}$ so yay.
The question is really good.
It's been eight years.