Let for a differentiable function f: ( 0, ∞ ) → R , f(x) - f(y) ≥ _...

Given: $f\left(x\right) - f\left(y\right) \ge \log_{e}\left(\dfrac{x}{y}\right) + x - y$

Now, taking $x = x + h \text{ and } y = x$ we get,

$f\left(x + h\right) - f\left(x\right) \ge \log_{e}\left(\dfrac{x + h}{x}\right) + x + h - x$

$f\left(x + h\right) - f\left(x\right) \ge \log_{e}\left(\dfrac{x + h}{x}\right) + h$

Now, using ${f}^{'}\left(x\right) = \lim_{h \to 0}\dfrac{f\left(x + h\right) - f\left(x\right)}{h}$

$⇒ {f}^{'}\left(x\right) = \lim_{h \to 0}\dfrac{\log_{e}\left(\dfrac{x + h}{x}\right) + h}{h}$

$⇒ {f}^{'}\left(x\right) = \lim_{h \to 0}\dfrac{\log_{e}\left(1 + \dfrac{h}{x}\right)}{x \cdot \left(\dfrac{h}{x}\right)} + 1$

$⇒ {f}^{'}\left(x\right) = \dfrac{1}{x} + 1$

$⇒ {f}^{'}\left(\dfrac{1}{{x}^{2}}\right) = {x}^{2} + 1$

$⇒ \mathop{∑}_{x = 1}^{20}{f}^{'}\left(\dfrac{1}{{x}^{2}}\right) = \mathop{∑}_{x = 1}^{20}\left({x}^{2} + 1\right)$

$⇒ \mathop{∑}_{x = 1}^{20}{f}^{'}\left(\dfrac{1}{{x}^{2}}\right) = \mathop{∑}_{x = 1}^{20}{x}^{2} + 20$

$⇒ \mathop{∑}_{x = 1}^{20}{f}^{'}\left(\dfrac{1}{{x}^{2}}\right) = \dfrac{20 \times 21 \times 41}{6} + 20$

$⇒ \mathop{∑}_{x = 1}^{20}{f}^{'}\left(\dfrac{1}{{x}^{2}}\right) = 2870 + 20$

$⇒ \mathop{∑}_{x = 1}^{20}{f}^{'}\left(\dfrac{1}{{x}^{2}}\right) = 2890$