Let a sequence ${[a_n]}$ be defined such that $a_1=1$ and when $n\ge1$, $a_{n+1}=a_n+\frac{1}{a_n}$. Then, show that:$12
Let a sequence ${[a_n]}$ be defined such that $a_1=1$ and when $n\ge1$, $$a_{n+1}=a_n+\frac{1}{a_n}$$
Then, show that:$$12<a_{75}<15$$
(the bracket in ${[a_n]}$ does not denote GIF, it is a general sequence)
All I got to know was that using the recursive relation we get:$$a_{75} = a_1+(\frac{1}{a_1} + \frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}+...+\frac{1}{a_{74}})$$
I do not know how to proceed further. When I tried to express everything in form of $a_1$, it is becoming too messy and complicated.
Like:
$$a_2 = a_1 + \frac{1}{a_1}$$
$$a_3 = a_1+\frac{1}{a_1} +\frac{1}{a_1 + \frac{1}{a_1}} $$
$$a_4 = a_1+\frac{1}{a_1} +\frac{1}{a_1 + \frac{1}{a_1}} + \frac{1}{a_1+\frac{1}{a_1} +\frac{1}{a_1 + \frac{1}{a_1}}}$$
$$a_5 = a_1+\frac{1}{a_1} +\frac{1}{a_1 + \frac{1}{a_1}} + \frac{1}{a_1+\frac{1}{a_1} +\frac{1}{a_1 + \frac{1}{a_1}}}+\frac{1}{a_1+\frac{1}{a_1} +\frac{1}{a_1 + \frac{1}{a_1}} + \frac{1}{a_1+\frac{1}{a_1} +\frac{1}{a_1 + \frac{1}{a_1}}}}$$
And so on...
Any hints or help would be appreciated!

This question has to be done without using induction.

Use
$$ a_n^2 + 3 \gt a_{n+1}^2 \gt a_n^2 + 2$$
