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Question: Let for a differentiable function $f: (0, \infty) \rightarrow R$, $f(x)-f(y) \geq \log_e \left(\fra...

Let for a differentiable function f:(0,∞)→Rf: (0, \infty) \rightarrow R,

f(x)−f(y)≥log⁡e(xy)+x−y,∀x,y∈(0,∞)f(x)-f(y) \geq \log_e \left(\frac{x}{y}\right) + x -y, \forall x, y \in (0, \infty).

Then ∑n=120f′(1n2)\sum_{n=1}^{20} f' \left(\frac{1}{n^2}\right) is equal to _______.

Answer

2890

Explanation

Solution

The given inequality implies f(x)−log⁡ex−x=Cf(x) - \log_e x - x = C. Differentiating gives f′(x)=1x+1f'(x) = \frac{1}{x} + 1. The sum is ∑n=120(n2+1)=∑n=120n2+∑n=1201=20(21)(41)6+20=2870+20=2890\sum_{n=1}^{20} (n^2+1) = \sum_{n=1}^{20} n^2 + \sum_{n=1}^{20} 1 = \frac{20(21)(41)}{6} + 20 = 2870 + 20 = 2890.