Solve the sequences inequality If $a_1=1$ and $a_n=a_{n-1}+\dfrac{1}{a_{n-1}}$ for $n≥2$ , then prove that $12 < a_{75} < 15$ ?
I have tried solving this by:
 $$a_{75} - a_1 = \frac{1}{a_1}+\frac{1}{a_2}....+\frac{1}{a_{74}}$$
 A: Let $b_n=a_n^2$. Then $b_1=1$ and:
$$ b_{n+1} = b_n + 2 + \frac{1}{b_n}, $$
hence $b_n\geq 2n-1$ holds by induction, so $a_{75}\geq \sqrt{149}>12$.
On the other hand, $b_n\geq 2n-1$ also gives:
$$ b_{n+1} \leq b_n + 2 + \frac{1}{2n-1}, $$
hence:
$$ b_n \leq (2n-1)+\sum_{k=1}^{n-1}\frac{1}{2k-1} $$
holds by induction and we get $a_{75}\leq 13$ as well.
A: Use the hint given by Did:
We have that $a_n^2=a_{n-1}^2+\frac{1}{a_{n-1}^2}+2$. Thus $a_n^2\geq a_{n-1}^2+2$ for all $n\geq 2$. Hence $a_n^2\geq 2\cdot (n-1) +1$. Hence $a_{75}\geq \sqrt{2\cdot 74+1}\geq \sqrt{2\cdot 72}=\sqrt{144}=12$.
For the other inequality, use the following reasoning:
$$a_n^2=a_{n-1}^2+\frac{1}{a_{n-1}^2}+2\leq a_{n-1}^2+\frac{1}{2(n-2)+1}+2.$$
Here we used the inequality $a_{n-1}^2\geq 2\cdot (n-2) +1$ in the denominator. Now replace $a_{n-1}^2$ by $a_{n-2}^2+\frac{1}{a_{n-2}^2}+2$ and again use the inequality in the denominator. Continue in this fashion to obtain the result. (Do this yourself, it requires a bit of work). 
All credits go to @Did for his brilliant hint!
A: Prove using induction that for all $n \geq 2$
$$ \sqrt{2n} \leq a_n \leq \sqrt{3n}.$$ 
