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

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:

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