Inequality $ 2\left(E_{k+1}^2+E_{k-1}^2\right)+5 E_k^2<1$ Consider the following inequality:
$$ 2\left(E_{k+1}^2+E_{k-1}^2\right)+5 E_k^2<1$$
where 
$$E_k=\frac{F_k}{2^k}$$
and $F_k$ Fibonacci number. 
I need some suggestions. Thank you!
 A: The OP presumably means the Fibonacci numbers indexed with $F_1=F_2=1$, so that $E_1={1\over2}$, $E_2=E_3={1\over4}$, $E_4={2\over16}$, etc.
With this in mind, note first that
$$E_k={F_k\over2^k}={F_{k-1}+F_{k-2}\over2^k}\le{2F_{k-1}\over2^k}={F_{k-1}\over2^{k-1}}=E_{k-1}$$
Consequently $E_k\lt{1\over3}$ for $k\gt1$, hence if $k\gt2$ we have
$$2(E_{k+1}^2+E_{k-1}^2)+5E_k^2\le9E_{k-1}^2\lt9\left({1\over3}\right)^2=1$$
Finally, to mop things up, if $k=2$ we have
$$2(E_3^2+E_1^2)+5E_2^2=2({1\over16}+{1\over4})+5\cdot{1\over16}={15\over16}\lt1$$
Note, however, that
$$2(E_2^2+E_0^2)+5E_1^2=2({1\over16}+0)+5\cdot{1\over4}={11\over8}\gt1$$
so the OP either didn't intend the inequality to start at $k=1$ or else intended to index the Fibonacci numbers with $F_1=0$, $F_2=1$.
A: Here is the gist of the proof by induction. 
Base Case. 
...
Inductive Step. Suppose for some j> something :), $$ 2\left(E_{j+1}^2+E_{j-1}^2\right)+5 E_j^2<1$$ 
So $$E_{j+2} = \frac{F_{j+2}}{2^{j+2}}=\frac{F_{j+1} + F_j}{2^{j+2}} = \frac{E_{j+1}}{2} + \frac{E_{j}}{4} = \frac{F_{j} + F_{j-1}}{2^{j+2}} + \frac{E_j}{4} = E_j/2 $$ 
...
And this is where you show $$ 2\left(E_{j+2}^2+E_{j}^2\right)+5 E_{j+1}^2<1$$
