Question

Let for a differentiable function \(f : (0, \infty) \rightarrow \mathbb{R}\), \(f(x) - f(y) \geq \log_e \left( \frac{x}{y} \right) + x - y, \quad \forall \; x, y \in (0, \infty).\) 
Then \(\sum_{n=1}^{20} f'\left(\frac{1}{n^2}\right)\) is equal to ____.

Answer 1

Correct Answer: 2890

We start by differentiating the inequality:
\(f(x) - f(y) \geq \log_e \left( \frac{x}{y} \right) + x - y\). 
Rewriting it as:
\(f(x) - \log_e(x) - x \geq f(y) - \log_e(y) - y\), this implies \(g(x)=f(x) - \log_e(x) - x\) is non-decreasing. Consequently, \(g'(x) \geq 0\) or \(f'(x) \geq \frac{1}{x} + 1\).
Now, compute the sum:
\(\sum_{n=1}^{20} f'\left(\frac{1}{n^2}\right) \geq \sum_{n=1}^{20} \left( n^2 + 1\right)\).
The sum becomes:
\(\sum_{n=1}^{20} \left(n^2 + 1\right) = \sum_{n=1}^{20} n^2 + \sum_{n=1}^{20} 1\)
Using \(\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}\), 
we get:
\(\sum_{n=1}^{20} n^2 = \frac{20 \times 21 \times 41}{6}=2870\)
Adding, \(\sum_{n=1}^{20} 1 = 20\), gives us:
\(\sum_{n=1}^{20} n^2 + \sum_{n=1}^{20} 1= 2870 + 20 = 2890\).
Thus, \(\sum_{n=1}^{20} f'\left(\frac{1}{n^2}\right) \geq 2890\), and falls within the range \([2890, 2890]\).

Answer 2

Given: \(f(x) - f(y) \geq \ln x - \ln y + x - y\)

Rewriting:

\(f(x) - f(y) = \frac{\sqrt{x} - \sqrt{y}}{x - y} \cdot \frac{\sqrt{x} + \sqrt{y}}{x - y}\)

- Case 1: Let \( x \to y \)
\(\lim_{y \to x^-} f' \left( \frac{1}{x} \right) \geq \frac{1}{x} + 1 \quad \text{...(1)}\)

- Case 2: Let \( x \neq y \)

 \(\lim_{y \to x^+} f' \left( x^+ \right) \leq \frac{1}{x} + 1 \quad \text{...(2)}\)
Thus, \(f'(x) = \frac{1}{x+1}\).

Now, substitute \( f' \left( \frac{1}{x} \right) = n + 1 \) into the sum:

\(\sum_{n=1}^{20} f' \left( \frac{1}{x} \right) = \sum_{n=1}^{20} \left( n^2 + 1 \right)\)
 

Calculating:

\(\sum_{n=1}^{20} n^2 = \frac{20 \cdot 21 \cdot 41}{6} = 2870, \quad \sum_{n=1}^{20} 1 = 20\)
 

Therefore:

\(\sum_{n=1}^{20} f' \left( \frac{1}{x} \right) = 2870 + 20 = 2890\)