Have no idea how to start in working on this problem.
The numbers $F_0,F_1,F_2,...$ are defined as follows :$$F_0 = 0, F_1 = 1, F_{n+2}=F_{n+1}+F_n$$ for $n = 0,1,2,...$. Prove that for any $n\geq 0$ we have$$F_n \leq \left(\frac{1+\sqrt5}{2}\right)^{n-1}$$
Use induction.
Notice the base case, $n=0$ and $n=1$ is true.
Now, assume that it is true for $n=m-1$ and $n=m$. Let $\phi=\frac{\sqrt{5}+1}{2}$. We have that $$F_{m-1} \le \phi^{m-2} \tag{1}$$ $$F_{m} \le \phi^{m-1} \tag{2}$$
Note that if $$\phi =\frac{\sqrt{5}+1}{2} \iff 2 \phi-1=\sqrt{5} \implies 4\phi^2-4 \phi-4=0 $$ So $\phi^2=\phi +1$. Multiplying each side by $\phi^{m-2}$ we get $$\phi^m=\phi^{m-1}+\phi^{m-2}$$ Now, adding $\text{(1)}$ and $\text{(2)}$ we have $$F_{m+1}=F_{m}+F_{m-1} \le \phi^{m-1}+\phi^{m-2}=\phi^{m}$$ So it is true for $n=m+1$. We are done.