Form the inequality from the following mathematical statement: The product of 16 and 3 is greater than the sum of 16 and 3

Step 1: Find the sum of $2\frac{1}{4}$ and $\frac{3}{4}$. First, convert the mixed number to an improper fraction: $2\frac{1}{4} = \frac{(2 \times 4) + 1}{4} = \frac{8 + 1}{4} = \frac{9}{4}$ Now, add the fractions: $\frac{9}{4} + \frac{3}{4} = \frac{9 + 3}{4} = \frac{12}{4}$ Simplify the sum: $\frac{12}{4} = 3$

Step 2: Convert $1\frac{1}{8}$ to an improper fraction. $1\frac{1}{8} = \frac{(1 \times 8) + 1}{8} = \frac{8 + 1}{8} = \frac{9}{8}$

Step 3: Subtract the result from Step 2 from the result of Step 1. We need to calculate $3 - \frac{9}{8}$. To subtract, find a common denominator. The number 3 can be written as $\frac{3}{1}$. The least common multiple of 1 and 8 is 8. Convert $3$ to an equivalent fraction with a denominator of 8: $3 = \frac{3 \times 8}{1 \times 8} = \frac{24}{8}$ Now perform the subtraction: $\frac{24}{8} - \frac{9}{8} = \frac{24 - 9}{8} = \frac{15}{8}$

Step 4: Convert the improper fraction to a mixed number. Divide 15 by 8: $15 \div 8 = 1 with a remainder of 7$ So, the mixed number is $1\frac{7}{8}$.

The final answer is $\boxed{1\frac{7}{8}}$.