Is it true that $\lceil (3/2)^n \rceil - (3/2)^n > (3/4)^n$ for all $n>1$? It is easy to check that
$$
\lceil (3/2)^n \rceil - (3/2)^n \ge 1/2^n
$$
(to see this, just look at the binary representation of $(3/2)^n$).
Prove or disprove the stronger statement that
$$
\lceil (3/2)^n \rceil - (3/2)^n > (3/4)^n
$$
for all integer $n>1$. Here $\lceil x \rceil$ is the ceiling function; e.g.
$\lceil 3/2 \rceil = 2$.
Note: Using PARI/GP, the statement has been verified up to $n=20000$. (For this verification, default precision may need to be increased and/or the original inequality should be restated in terms of the fractional part $\{(3/2)^n\}$, as Gottfried Helms explains below.)
 A: From $3^n=2^n\cdot k+ r, 0\leq r<2^n-1$
$$\left \lceil \frac{3^n}{2^n} \right \rceil - \frac{3^n}{2^n} > \frac{3^n}{2^{2n}} \iff 
2^n\left \lceil \frac{3^n}{2^n} \right \rceil - 3^n > \frac{3^n}{2^n} \iff \\
2^n\left(\left \lfloor \frac{3^n}{2^n}\right \rfloor+1\right)-3^n>\frac{3^n}{2^n} \iff 
2^n(k+1)-3^n> \frac{3^n}{2^n}  \iff ...$$
but $3^n+2^n=2^n\cdot(k+1)+r$, thus
$$...\iff 2^n-r > \frac{3^n}{2^n} \tag{1}$$
It is worth noting that $r=3^n-2^n\left \lfloor \frac{3^n}{2^n}\right \rfloor$ and it is known that
$$r>2^n-\left \lfloor \frac{3^n}{2^n}\right \rfloor-2$$
admits only finitely many solutions for $n$ (although, none known). I.e. from some large enough $n_0$ onwards 
$$r\leq 2^n-\left \lfloor \frac{3^n}{2^n}\right \rfloor-2 \Rightarrow
2^n-r\geq 2^n - 2^n+\left \lfloor \frac{3^n}{2^n}\right \rfloor+2=\\
\left \lfloor \frac{3^n}{2^n}\right \rfloor+2 > \frac{3^n}{2^n}$$
and, via the equivalences from $(1)$, we have that the original inequality is valid, from some $n_0$ onwards.
A: This is not meant as an answer, just as an extended comment 
Rewriting the OP's formula $ \lceil(3/2)^N\rceil-(3/2)^N \gt (3/4)^N $ as
$$ 1 - (3/4)^N \gt \lbrace (3/2)^N \rbrace \tag 1
$$
as I know it from the formula in mathworld on fractional power parts (and its reference to the (open) Waring's problem) and even better as 
$$ 1  \gt \lbrace (3/2)^N \rbrace +  (3/4)^N \tag 2
$$
gives the idea to simply compare the leading bitstrings of $a(N)=\lbrace (3/2)^N \rbrace$ and $b(N)=(3/4)^N $ .         
Of course, the length of leading zero-bitstring of the latter ($b(N)$), let's call it $g(N)$, is easily computable. And simply, the length of the leading one-bitstring in $a(N)=\lbrace (3/2)^N \rbrace$ , let's call it $f(N)$, should be smaller than the length of the zero-bitstring to keep eq (2) true.             
So I looked at the lengthes of the leading one-bitstrings as function of $N$. This is a plot of that lengthes:

We see, that the leading bitstring of zeros in $b(N)$ grows so fast that there is no chance that its sum with the leading bitstring in $a(N)$ could ever result in becoming larger than $1$. The upper hullcurve of the function $f(N)$ has only logarithmic growth (while $g(N)$ grows linearly). It's really easy to conjecture from this that there are no counterexamples for eq (1) above  $N>7$ ...          
However, I've not yet a good idea to describe a pattern for $f(N)$ to get this all more handy; but from pictures like this I think, there should be a qualitative much better function to upper-bound the $\lbrace (3/2)^N \rbrace$ than this $(3/4)^N$-function...              

Just for comparision with that concept of length-of-leading-bitstring one can look at the atanh() function-inequality 
$$ \text{atanh}(2\cdot(1-(3/4)^N)-1) \gt \text{atanh}(2\cdot\{(3/2)^N\}-1) \gt \text{atanh}(2\cdot(3/4)^N-1) \tag 3 $$ (for $N>7$, built after the mathworld-reference to the Waring-problem)

The red-painted upper and lower hullcurves for the fractional part of $(3/2)^N$ behave roughly logarithmical with $N$ while again, the curves for the boundaries of eq (3) go roughly linear with $N$.        

Unfortunately, still no proof for that shape of $\{(3/2)^N\}$ seems to be present, so this is all only suggestion and only likelihood deserving the final step of proving to be valid without exceptions for $N \gt n_0=7$ .
