Let $\{ a_n\}_{n \ge 0}$ be a sequence of real numbers such that $a_{n+1} \ge a_n^2 + \frac{1}{5}$ for all $n \ge 0$. Prove that $\sqrt{a_{n+5}} \ge a_{n-5}^2$ for all $n \ge 5$.
Problem
Source: USA TST 2001 #1
Tags: inequalities, inequalities unsolved
darij grinberg
18.11.2005 16:56
This trivially follows from $a_{n+5}\geq a_n^2$ and $a_n\geq a_{n-5}^2$, what, in turn, follows from the inequality proved at http://www.mathlinks.ro/Forum/viewtopic.php?t=28048 . darij
April
20.05.2007 17:54
We have: $\left\{\begin{array}{c}a_{n+1}\ge a_{n}^{2}+\frac{1}{5}\\ a_{n+2}\ge a_{n+1}^{2}+\frac{1}{5}\\ a_{n+3}\ge a_{n+2}^{2}+\frac{1}{5}\\ a_{n+4}\ge a_{n+3}^{2}+\frac{1}{5}\\ a_{n+5}\ge a_{n+4}^{2}+\frac{1}{5}\end{array}\right\|$ $\Longrightarrow a_{n+5}\ge a_{n}^{2}+\left(a_{n+1}^{2}-a_{n+1}+\frac{1}{4}\right)+\left(a_{n+2}^{2}-a_{n+2}+\frac{1}{4}\right)$ $+\left(a_{n+3}^{2}-a_{n+3}+\frac{1}{4}\right)+\left(a_{n+4}^{2}-a_{n+4}+\frac{1}{4}\right) \Longrightarrow a_{n+5}\ge a_{n}^{2}$ Similar: $a_{n}\ge a_{n-5}^{2}$ Thus, $a_{n+5}\ge a_{n-5}^{4}\Longrightarrow\sqrt{a_{n+5}}\ge a_{n-5}^{2}\quad\mathbb{QED}$
AopsUser101
22.05.2020 03:59
We have the following:
$$a_{n+1} \ge a_n^2 + \frac{1}{5}$$$$a_{n+2} \ge a_{n+1}^2 + \frac{1}{5}$$$$a_{n+3} \ge a_{n+2}^2 + \frac{1}{5}$$$$a_{n+4} \ge a_{n+3}^2 + \frac{1}{5}$$$$a_{n+5} \ge a_{n+4}^2 + \frac{1}{5}$$Adding, we see that $a_{n+5}^2 - a_{n}^2 \ge \left( a_{n+1}-\frac12\right)^2+\left( a_{n+2}-\frac12\right)^2 +\left( a_{n+3}-\frac12\right)^2 + \left( a_{n+4}-\frac12\right)^2 \ge 0$. It follows that $a_{n} \ge a_{n-5}^2 \Longleftrightarrow a_{n+5} \ge a_{n-5}^4$.