Prove that `log_n(n+1)>log_(n+1)(n+2)` for any natural number `n > 1.`

Answer

To prove that \( \log_n(n+1) > \log_{n+1}(n+2) \) for any natural number \( n > 1 \), we can follow these steps:

### Step 1: Rewrite the logarithmic expressions
Using the change of base formula for logarithms, we can express both logarithms in terms of natural logarithms (or any common base):

\[ \log_n(n+1) = \frac{\log(n+1)}{\log(n)} \]
\[ \log_{n+1}(n+2) = \frac{\log(n+2)}{\log(n+1)} \]

### Step 2: Set up the inequality
We need to prove that:

\[ \frac{\log(n+1)}{\log(n)} > \frac{\log(n+2)}{\log(n+1)} \]

### Step 3: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:

\[ \log(n+1) \cdot \log(n+1) > \log(n) \cdot \log(n+2) \]

This simplifies to:

\[ (\log(n+1))^2 > \log(n) \cdot \log(n+2) \]

### Step 4: Analyze the inequality
To analyze this inequality, we can define a function \( f(x) = \log(x) \). The function \( f(x) \) is concave for \( x > 0 \), which means that the average of the logarithms is greater than or equal to the logarithm of the average:

\[ \frac{\log(n) + \log(n+2)}{2} \leq \log\left(\frac{n + (n+2)}{2}\right) = \log(n + 1) \]

### Step 5: Apply the concavity of logarithm
From the concavity property, we can deduce:

\[ \log(n+1) > \frac{\log(n) + \log(n+2)}{2} \]

Squaring both sides (since both sides are positive for \( n > 1 \)) gives:

\[ (\log(n+1))^2 > \frac{(\log(n) + \log(n+2))^2}{4} \]

### Step 6: Show that the right-hand side is less than the left-hand side
By expanding the right-hand side and using properties of logarithms, we can show that:

\[ \log(n) \cdot \log(n+2) < (\log(n+1))^2 \]

This completes the proof that:

\[ \log_n(n+1) > \log_{n+1}(n+2) \]

### Conclusion
Thus, we have proved that \( \log_n(n+1) > \log_{n+1}(n+2) \) for any natural number \( n > 1 \).

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