For a positive real `x(x gt 1)`, which one of the following correct?

Options

A. `x/(x+1) lt x lt log (x+1)`
B. `x lt x/(1+x) lt log (x+1)`
C. `log(x +1) lt x/(1+x) lt x`
D. `x/(1+x) lt log (x+1) lt x`

Answer

To solve the problem, we need to analyze the expressions given for a positive real number \( x \) where \( x > 1 \). The expressions we need to compare are:

1. \( \log(x + 1) \)
2. \( x \)
3. \( \frac{x}{x + 1} \)

### Step 1: Evaluate each expression for a specific value of \( x \)

Let's take \( x = 2 \) as an example:

- **Calculate \( \log(2 + 1) \)**:
\[ \log(3) \approx 0.4771 \]

- **Calculate \( x \)**:
\[ x = 2 \]

- **Calculate \( \frac{x}{x + 1} \)**:
\[ \frac{2}{2 + 1} = \frac{2}{3} \approx 0.6667 \]

### Step 2: Compare the values

Now we have the following values:

- \( \log(3) \approx 0.4771 \)
- \( x = 2 \)
- \( \frac{x}{x + 1} \approx 0.6667 \)

### Step 3: Arrange the values in increasing order

From the calculated values, we can arrange them as follows:

1. \( \log(3) \approx 0.4771 \) (smallest)
2. \( \frac{2}{3} \approx 0.6667 \)
3. \( 2 \) (largest)

Thus, the order is:
\[ \log(x + 1) < \frac{x}{x + 1} < x \]

### Step 4: Generalize the findings

For any \( x > 1 \):
- \( \log(x + 1) \) will always be less than \( \frac{x}{x + 1} \) because \( \log(x + 1) \) grows slower than the linear function \( x \).
- \( \frac{x}{x + 1} \) approaches 1 as \( x \) increases, but it will always be less than \( x \).

### Conclusion

Thus, the correct order for \( x > 1 \) is:
\[ \log(x + 1) < \frac{x}{x + 1} < x \]