Is my approach to proving by induction correct? I'm currently studying proof by induction by solving some exercises. I'd like to know whether my proof is correct, and by extension whether my approach is right.
Given a sequence $\left ( a_{n}\right ) ,n\in\mathbb{N},$
$$a_{1}=0,$$
$$a_{n+1}=\frac{a_{n}^{2}+6}{5},$$
prove that it is increasing and bounded above by $2$ using mathematical induction.
Proof that it is increasing:
$$P\left ( n\right ) :=a_{n+1}\geq a_{n},$$
$$P\left ( 1 \right )\equiv \frac{0^{2}+6}{5}\geq 0\equiv \frac{6}{5}\geq 0\equiv \top,$$
$$P\left ( k \right )\equiv \frac{a_{k}^{2}+6}{5}\geq a_k,$$
$$P\left ( k+1 \right )\equiv \frac{a_{k+1}^{2}+6}{5}\geq a_{k+1}\equiv \frac{\left ( \frac{a_{k}^{2}+6}{5} \right )^{2}+6}{5}\geq \frac{a_{k}^{2}+6}{5}$$
$$\equiv \left ( \frac{a_{k}^{2}+6}{5} \right )^{2}+6\geq a_{k}^{2}+6$$
$$\equiv\left ( \frac{a_{k}^{2}+6}{5} \right )^{2}\geq a_{k}^{2}$$
$$\equiv\frac{a_{k}^{2}+6}{5}\geq a_{k}\equiv P\left ( k \right ).$$
Proof that it is bounded above by $2$:
$$P\left ( n \right ):= a_{n}\leq 2,$$
$$P\left ( 1 \right )\equiv 0\leq 2\equiv \top ,$$
$$P\left ( k \right )\equiv \frac{a_{k}^{2}+6}{5}\leq 2,$$
$$P\left ( k+1 \right )\equiv \frac{a_{k+1}^{2}+6}{5}\leq 2\equiv \frac{{\left ( \frac{a_{k}^{2}+6}{5} \right )}^{2}+6}{5}\leq 2$$
$$\equiv {\left ( \frac{a_{k}^{2}+6}{5} \right )}^{2}+6\leq 10$$
$$\equiv {\left ( \frac{a_{k}^{2}+6}{5} \right )}^{2}\leq 4$$
$$\equiv \frac{a_{k}^{2}+6}{5}\leq 2\equiv P\left ( k \right )$$
 A: Disclaimer:
I'm going to be quite nitpicky with your proof. It's for your own benefit. In practice, most people won't care about some of these things that I'll point out but I think it's important for you to know. I'll start with general comments. Then, I'll comment on your argument and how you could make it simpler.

$P(n) := a_{n+1} \geq a_n$.

$P(n)$ is supposed to refer to an open sentence concerning the natural numbers $n \in \mathbb{N}$. So, you should write $P(n) :\iff a_{n+1} \geq a_n$.
Next thing; you are writing a proof. A proof is an argument and should read more like an essay than a string of symbols. You're supposed to include words explaining what you're doing. When you're trying to show that the base case holds, you should write:

Let's show that the base case holds. Let $n = 1$. Then, the following holds:
$$a_2 =\frac{a_1^2+6}{5} = \frac{0+6}{5} = \frac{6}{5} \geq 0 = a_1$$
This proves the base case.

Now, this is pretty simple so you might argue that you didn't need to write all of that. In fact, what you wrote is quite sufficient. But the real issues arise in the next step. Here, you assume that the result holds for some arbitrary $k \in \mathbb{N}$ and then you want to show it for the case when $n = k+1$. Once again, you should really write these things in plain English.
Now, those were the general comments on the style in which you've written your argument. As far as I can see, what you've written for showing that the sequence is non-decreasing is mostly correct. There's that little bit there where you've used the implication:
$$X^2 \geq Y^2 \implies X \geq Y$$
However, that's actually fine in this case because this sequence is non-negative (you can show this separately by induction). Here's a simple way to write out this argument.

We want to show that for all $n \in \mathbb{N}$, $a_{n+1} \geq a_n$. The base case is obvious. Now, assume that the result holds for some arbitrary but fixed $n \in \mathbb{N}$. Then:$$a_{n+1} = \frac{a_n^2+6}{5} \leq \frac{a_{n+1}^2+6}{5} = a_{n+2}$$
where that second inequality is justified because we assumed that $a_n \leq a_{n+1}$. It follows that $(a_n)_{n \in \mathbb{N}}$ is a non-decreasing sequence.

Next, let me do the same thing with the boundedness part of the result.

We want to show that $a_n \leq 2$ for all $n \in \mathbb{N}$. Indeed, this is true when $n = 1$. Assume that it is true for some arbitrary but fixed $n \in \mathbb{N}$. Then:
$$a_{n+1} = \frac{a_n^2+6}{5} \leq \frac{2^2+6}{5} = \frac{10}{5} = 2$$
and that proves the desired result. So, it follows by induction that $a_n \leq 2$ for all $n \in \mathbb{N}$.

Notice how everything I've written above flows like an argument, where you can clearly see what I'm doing at each step and where I'm using the induction hypothesis? It's all well and good to just work with symbols & there are situations where you can JUST use symbols in your proof without any words. That's not something I'd recommend that you do.
