How to prove this inequality: $f(2h-1)≤\frac{3h-1}{2}$ 
Let $I$ denote the set of odd integers. If $k \in I$, then $3k+1$ is even, so $3k+1=2^a k'$ with $k' \in I$ and $a \ge 1$. The Syracuse function is the function $f:I \to I$, for which $f(k) = k'$ (sequence A075677 in the OEIS).
Some properties of the Syracuse function are:
  
  
*
  
*$\forall k \in I, f(4k+1) = f(k)$. (Because $3(4k+1) + 1 = 12k+4 = 4(3k+1)$.)
  
*In more generality: For all $p \ge 1$ and $g \in I$, $f^{p-1}\left(2^ph-1\right) = 2 \cdot 3^{p-1}h - 1$. (Here, $f^{p-1}$ is the function iteration notation.)
  
*$$\forall h\in I, f(2h-1) \le \frac{3h-1}{2}$$
  
  
  The Collatz conjecture is equivalent to the statement that, $\forall k \in I$, there exists and integer $n \ge 1$ such that $f^n(k)=1$.

I dont understand why it must be
$$f(2h-1)≤\frac{3h-1}{2}??$$  
I cannot prove this inequality.

Consider $f(\color{blue}{2h-1})$; how many factors of two can we divide $3(\color{blue}{2h-1})+1$ by ?
\begin{eqnarray*}
3(2h-1)+1=6h-2=2(3h-1)
\end{eqnarray*}
and $3h-1$ is even, so there is at least another factor of $2$. So 
\begin{eqnarray*}
f(2h-1) \leq \frac{3h-1}{2}.
\end{eqnarray*}
