Let us define a series $\{a(n)\}$ such that $a(n)= a(n-1) + \frac{1}{a(n-1)}$

Notice first that

$$ a_n^2 = a_{n-1}^2 + 2 + \frac{1}{a_{n-1}^2} > a_{n-1}^2 + 2. \tag{*}$$

Recursively applying this inequality and using the fact that $a_2 = 2$,

$$ \forall n \geq 2, \qquad a_n^2 \geq 2n \tag{1} $$

Now plugging this inequality back to $\text{(*)}$, for $n \geq 3$ we have

$$ a_n^2 \leq a_{n-1}^2 + 2 + \frac{1}{2(n-1)} $$

with strict inequality when $n \geq 3$. Recursively applying this inequality yields

$$ a_n^2 \leq a_2^2 + \sum_{k=2}^{n-1} \left(2 + \frac{1}{2k} \right) = 2n + \sum_{k=2}^{n-1} \frac{1}{2k} \tag{2} $$

When $n = 75$, we obtain a lower bound from $\text{(1)}$:

$$ 12^2 < 150 \leq a_{75}^2 $$

and an upper bound from $\text{(2)}$:

$$ a_{75}^2 \leq 150 + \sum_{k=2}^{74} \frac{1}{2k} < 150 + \frac{\log 74}{2} < 150 + \frac{\log_2 128}{2} < 169 = 13^2. $$

This proves that $ 12 < a_{75} < 13 $ and hence the claim follows.