Would you please let me know how do I improve this proof?
The numbers $$F_{0},F_{1},F_{2},...$$ are defined as follows (this is a definition by mathematical induction, by the way): $$F_{0} = 0, F_{1} = 1, F_{n+2} = F_{n+1} + F_{n} \ for \ n=0,1,2,...$$
Prove that for any $$n≥0$$ we have
$$F_{n} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{n−1}$$
proof
Let P(n) be the proposition that
$$F_{n} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{n−1}$$
And we know
$$F_{0} = 0, F_{1} = 1, F_{n+2} = F_{n+1} + F_{n} \ for \ n=0,1,2,...$$
Then set up the base cases P(0),P(1),P(2) as
$$F_{0} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{0−1},F_{1} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{1−1},F_{2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{2−1}$$
we know from the fibonacci sequence and the given information, $$F_{0} = 0,F_{1} = 1,F_{2} = 1$$
Therefore, P(0),p(1),p(2) is true.
Assume that there is an $$k, (k \in \mathbf{Z})$$, then p(k) is
$$F_{k} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k−1}$$
And because $$F_{n+2}>F_{n+1}>F_{n} \ and \ F_{n+2} = F_{n+1} + F_{n} \ for \ n=0,1,2,...$$
And then by PMI, $$p(k) \implies p(k+1) \implies p(k+2)$$ so that
$$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k+2−1}=\left(\frac{1 + \sqrt {5}}{2}\right)^{k+1}$$ and because
$$F_{k+2} ＝ F_{k+1} ＋ F_{k}$$
Therefore $$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}+\left(\frac{1 + \sqrt {5}}{2}\right)^{k-1}$$
Therefore $$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}+\left(\frac{1 + \sqrt {5}}{2}\right)^{k}\left(\frac{1 + \sqrt {5}}{2}\right)^{-1}$$
$$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}\left(1+\left(\frac{1 + \sqrt {5}}{2}\right)^{-1}\right)$$
$$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}\left(1+\frac{2}{1 + \sqrt {5}}\right)$$
$$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}\left(\frac{1 + \sqrt {5}}{1 + \sqrt {5}}+\frac{2}{1 + \sqrt {5}}\right)$$
$$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}\left(\frac{3+ \sqrt {5}}{1 + \sqrt {5}}\right)$$
because $$\left(\frac{3+ \sqrt {5}}{1 + \sqrt {5}}\right)=\left(\frac{1 + \sqrt {5}}{2}\right)$$
Therefore $$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k+1}$$
In conclusion, $$p(k) \implies p(k+1) \implies p(k+2)$$ is true.
By PMI, $$p(n) \implies p(n+1) \implies p(n+2) \ and p(n)$$ is true.
Base cases are fine.
At the inductive hypothesis you must assume that $P(k)$ and $P(k-1)$ are true. You have only said to assume $P(k)$
You could use "Strong induction" and assume that for all $i\le k, P(i)$ is true.
And then you seem to spin a while, to get to the point.
Show that $P(k+1)$ is true based on the assumption $P(k)$ and $P(k-1)$ are true
let $\phi = \frac {1 + \sqrt 5}{2}$
Show that
$F_{k-1} < \phi^{k-2} , F_{k} < \phi^{k-1} \implies F_{k+1}<\phi^{k}$
$F_{k+1} = F_k + F_{k-1}$
$F_k + F_{k-1}<\phi^{k-1} + \phi^{k-2}$
$F_{k+1}<\phi^{k-2} (\phi+1)$
I say $\phi^2 = \phi+1$
$\left(\frac {1+\sqrt 5}{2}\right)^2 = \frac {6+2\sqrt 5}{4} = 1+\frac {1+\sqrt 5}{2}$
$F_{k+1}<\phi^{k}$
QED