How can i prove that $a_n \geq a_{n+1}$ for $n > 1$?

Question

given $a_{n+1}= \frac{1}{2}(a_n+\frac{5}{a_n})$ , $a_1=1$

How can i prove that $a_n \geq a_{n+1}$ for $n > 1$?

I tried by induction, but i don't know how to do the stage of the "prove" in the induction.

Answer 1

$$a_{n+1}= \frac{1}{2}(a_n+\frac{5}{a_n})$$

we have, $a_n\ge \sqrt5$ for $n\ge 2.$(why?) indeed, By AM-GM inequality

$$a_{n+1}= \frac{1}{2}(a_n+\frac{5}{a_n}) \ge \sqrt{a_n*\frac{5}{a_n}}\ge \sqrt5$$

Now consider, $$f(x) = \frac{x}{2}- \frac{5}{2x}\implies f'(x) = \frac12+ \frac{5}{2x^2}>0$$ so f is increasing, Then for every $x> \sqrt{5}$ we have, $$f(x) >f(\sqrt5)=0$$

Therefore, we have $$ f(a_n)= \frac{a_n}{2}- \frac{5}{2a_n}\ge 0\Longleftrightarrow a_n\ge a_{n+1}$$ which is true.

Answer 2

Attempt:

$a_n > 0,$ for all $n\ge1$.

AM-GM: $a_{n+1} \ge √5.$

(Recall: $(1/2)(a_n+5/a_n) \ge $

$\sqrt{a_n×(5/a_n)} =√5)$

Note: $a_1=1;$ $a_n \ge √5$ for $n\ge 2.$

For $n \ge 2:$

$a_{n+1}-a_{n} = (1/2)(5/a_n -a_n) \le (1/2)(5/(√5) -a_n)=$

$(1/2)(√5-a_n) \le 0,$ since $a_ n \ge √5.$

Hence : $a_{n+1} \le a_n$.