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

The inequality is differentiated as: \(f(x) - f(y) \geq \log_e \left( \frac{x}{y} \right) + x - y\). This is rewritten as: \(f(x) - \log_e(x) - x \geq f(y) - \log_e(y) - y\), indicating that \(g(x)=f(x) - \log_e(x) - x\) is non-decreasing. Therefore, \(g'(x) \geq 0\), which means \(f'(x) \geq \frac{1}{x} + 1\). The sum is computed as: \(\sum_{n=1}^{20} f'\left(\frac{1}{n^2}\right) \geq \sum_{n=1}^{20} \left( n^2 + 1\right)\). The sum \(\sum_{n=1}^{20} \left(n^2 + 1\right)\) is expanded to \(\sum_{n=1}^{20} n^2 + \sum_{n=1}^{20} 1\). Using the formula \(\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}\), we find \(\sum_{n=1}^{20} n^2 = \frac{20 \times 21 \times 41}{6}=2870\). Adding \(\sum_{n=1}^{20} 1 = 20\), we get \(2870 + 20 = 2890\). Consequently, \(\sum_{n=1}^{20} f'\left(\frac{1}{n^2}\right) \geq 2890\), with the range being \([2890, 2890]\).