Just wanted to ask if someone could help me with this proof: $f_{n+1}<(\frac{7}{4})^{n}, n\geq 1$, where $f_{n+1} = f_{n}+f_{n-1}.$ The base case where $n=1$ is straightforward. I wrote down an induction hypothesis:
$f_{k+1}<(\frac{7}{4})^{k},$ for some $k\geq 1.$
Induction step: By induction this implies that: $f_{k+2}<(\frac{7}{4})^{(k+1)}.$
I started the proof with the induction hypothesis: $f_{k+1}<(\frac{7}{4})^{k},$ and deduced from there that: $f_{k+1}+f_{k}<(\frac{7}{4})^{k}+f_{k},$ which is: $f_{k+2}<(\frac{7}{4})^{k}+f_{k}.$ (By definition of a Fibonacci number.)
I then tried to show that if: $(\frac{7}{4})^{k}+f_{k}<(\frac{7}{4})^{(k+1)},$ then this must imply that: $f_{k+2}<(\frac{7}{4})^{(k+1)}.$