5 /leq 4x + 1 < 25

To solve the compound inequality \( \frac{5}{4} \leq 4x + 1 < 25 \), we can break it down into two parts and solve each part step by step.

### Part 1: Solve \( \frac{5}{4} \leq 4x + 1 \)

1. Subtract 1 from both sides:
\[
\frac{5}{4} - 1 \leq 4x
\]

2. Convert 1 into a fraction with a common denominator:
\[
\frac{5}{4} - \frac{4}{4} = \frac{1}{4}
\]
So we have:
\[
\frac{1}{4} \leq 4x
\]

3. Divide both sides by 4:
\[
\frac{1}{16} \leq x
\]

### Part 2: Solve \( 4x + 1 < 25 \)

1. Subtract 1 from both sides:
\[
4x < 25 - 1
\]
\[
4x < 24
\]

2. Divide both sides by 4:
\[
x < 6
\]

### Combine the results

From the parts we solved, we have:
\[
\frac{1}{16} \leq x < 6
\]

### Final Answer
The solution to the compound inequality is:
\[
\frac{1}{16} \leq x < 6
\]

In interval notation, this is:
\[
\left[ \frac{1}{16}, 6 \right)
\]