Strong mathematical induction proof Can someone please work me through this problem?

Let $(f_0, f_1,f_2,...)$ be the Fibonacci sequence, that is, $f_0=0$, $f_1=1$, and $f_n = f_{n−2} + f_{n−1}$ for all $n\ge2$. Prove that $f_n>(5/4)^n$ for all integers $n\ge3$.

I appreciate any kind of help :))
 A: Upon working, I realised the normal single assumption in the induction hypothesis won't do this. 
Define $P(n)$ as the statement $f_n > (5/4)^n \text{ and } f_{n+1} > (5/4)^{n+1}$, only being true if both conditions are true. Prove the base case for $n = 3$ accordingly, i.e, verify $2>(5/4)^3$ and $3>(5/4)^4$.
Assume $P(k)$ holds for some natural number $k\ge3$. This implies:
$$f_k>(5/4)^k, \space f_{k+1} > (5/4)^{k+1}$$ For the induction step, it will be sufficient to show $f_{k+2} > (5/4)^{k+2}$, since it implies $P(k+1)$ is true, as we already have $f_{k+1} > (5/4)^{k+1}$. We know
$$f_{k+2} = f_{k+1} + f_k$$ By induction hypothesis, $f_k> (5/4)^k$ and $f_{k+1}>(5/4)^{k+1}$. We get
$$f_{k+2} = f_k+f_{k+1}> (5/4)^k + (5/4)^{k+1}$$
Simplifying RHS by factoring $(5/4)^k$:
$$f_{k+2} > \left(\frac54\right)^k\times \frac{9}{4}>\left(\frac54\right)^k \times \left(\frac54\right)^2$$
Thus,
$$\color{green}{f_{k+2} > \left(\frac54\right)^{k+2}}$$
Which completes our induction step. I leave the rest to you.
