Question

Let for a differentiable function f: (0, ∞) → R,

$f(x)-f(y) \ge log_e(\frac{x}{y})+x-y, \forall x, y \in (0, \infty).$

Then

$\sum_{n=1}^{20} f'(\frac{1}{n^2})$ is equal to ____.

Answer 1

2890

Answer 2

Given the inequality $f(x)-f(y) \ge \log_e(\frac{x}{y})+x-y, \forall x, y \in (0, \infty)$. Rearrange the inequality: $f(x)-f(y) - (x-y) - (\log_e x - \log_e y) \ge 0$ Consider a new function $g(x) = f(x) - x - \log_e x$. The inequality can be rewritten as $g(x) - g(y) \ge 0$.

Case 1: Let $x > y$. Then $g(x) \ge g(y)$, which implies $\frac{g(x)-g(y)}{x-y} \ge 0$. Taking the limit as $x \to y^+$: $g'(y) \ge 0$.

Case 2: Let $x < y$. Then $g(x) \le g(y)$, which implies $\frac{g(x)-g(y)}{x-y} \le 0$ (since $x-y < 0$, the inequality sign flips). Taking the limit as $x \to y^-$: $g'(y) \le 0$.

Since $f$ is differentiable, $g$ is also differentiable. For $g'(y)$ to exist, it must satisfy both $g'(y) \ge 0$ and $g'(y) \le 0$. This implies $g'(y) = 0$.

Now, let's find $g'(x)$: $g'(x) = \frac{d}{dx}(f(x) - x - \log_e x) = f'(x) - 1 - \frac{1}{x}$. Setting $g'(x) = 0$: $f'(x) - 1 - \frac{1}{x} = 0$ $f'(x) = 1 + \frac{1}{x}$.

We need to calculate $\sum_{n=1}^{20} f'(\frac{1}{n^2})$. Substitute $x = \frac{1}{n^2}$ into $f'(x)$: $f'(\frac{1}{n^2}) = 1 + \frac{1}{\frac{1}{n^2}} = 1 + n^2$.

Now, calculate the sum:

$\sum_{n=1}^{20} f'(\frac{1}{n^2}) = \sum_{n=1}^{20} (1 + n^2) = \sum_{n=1}^{20} 1 + \sum_{n=1}^{20} n^2 = 20 \times 1 + \frac{20(20+1)(2 \times 20+1)}{6} = 20 + \frac{20 \times 21 \times 41}{6} = 20 + 10 \times 7 \times 41 = 20 + 2870 = 2890$