Let the sequence $\{a_n\}$ be defined by $a_{n+1} = \frac14(a_n^2 +3)$. We want to prove that if the first term $a_1$ is between $0$ and $1$ then the sequence converges.
My question is why do we have to use induction twice. First by induction we show $a_n$ is between $0$ and $1$, but then induction is used again to show that the sequence is increasing and hence converges by MCT. But can't we deduce immediately that since $$a_{n+1}-a_n=\frac14 (a_{n}-1)( a_{n}-3 )$$ and since $a_n$ is between $0$ and $1$ for all $n$ hence $a_{n+1}-a_n$ is greater than $0$ for all $n$ and hence increasing? Why do all books use induction in the second part also?
Observe that: $0<a_{n+1} = \dfrac{a_n^2+3}{4} < \dfrac{1+3}{4} = 1$ by induction. And you proved that $\{a_n\}$ is increasing by induction as well. Thus it converges to $L$ such that: $L = \dfrac{L^2+3}{4} \to L^2-4L+3=0\to L = 1,3$. But we rule out $L=3$ since $L\leq 1$. So $L = 1$.
Note: To prove the sequence increases there are two ways: by using boundedness, or by using induction. Induction seems to be more popular. Here is how:
$a_{n+1}-a_n = \dfrac{a_n^2+3}{4}-\dfrac{a_{n-1}^2+3}{4}=\dfrac{(a_n+a_{n-1})(a_n-a_{n-1})}{4}>0$ since $a_n-a_{n-1} > 0$ by inductive step.