Proof that a subtree of a red-black tree has no more than $\frac{3n}{4}$ nodes
Here is a natural approach to explore the extreme case. 

Let node $r$ be the root of a red-black tree. 

Let its left subtree be a red-black tree of black height $h$ that has the minimum number of the nodes among red-black tree of the same black height. 

Let its right subtree be a red-black tree of the same black height $h$ that has the maximum number of the nodes among red-black tree of the same black height.

Now let $h=1,2,3,\cdots$ be larger and larger. Can you see the trend?





Here is a related exercise.

Exercise (red-black trees may not be weight-balanced). Given $\alpha\lt 1$, show that there is a red-black tree which has a proper subtree that has more than $\alpha$ of all nodes.