Solution
= Solution
Use <mathematical induction> with two consecutive <base cases>: $F_2=1=(3/2)^0$ and $F_3=2\geq(3/2)^1$. Suppose the bounds hold for exponents $n$ and $n+1$. The <Fibonacci recurrence> gives
$$
F_{n+4}=F_{n+3}+F_{n+2}\geq(3/2)^{n+1}+(3/2)^n=\frac52(3/2)^n\geq\frac94(3/2)^n=(3/2)^{n+2}.
$$
Thus the consecutive pair of bounds advances one step. \b[$F_{n+2}\geq(3/2)^n$ for every $n\geq0$.]