Let for a differentiable function f : ( 0 , ∞ ) → R , f x - f y ≥ log e x y + x - y , ∀ x , y ∈ 0 , ∞ . Then…

Given: $f\left(x\right)-f\left(y\right)\ge {\log }_e\left(\frac{x}{y}\right)+x-y$

Now, taking $x=x+h \& y=x$ we get,

$f\left(x+h\right)-f\left(x\right)\ge {\log }_e\left(\frac{x+h}{x}\right)+x+h-x$

$f\left(x+h\right)-f\left(x\right)\ge {\log }_e\left(\frac{x+h}{x}\right)+h$

Now, using $f^{\text{'}}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

$\Rightarrow f^{\text{'}}\left(x\right)=\underset{h\to 0}{\lim }\frac{{\log }_e\left(\frac{x+h}{x}\right)+h}{h}$

$\Rightarrow f^{\text{'}}\left(x\right)=\underset{h\to 0}{\lim }\frac{{\log }_e\left(1+\frac{h}{x}\right)}{x·\left({\displaystyle \frac{h}{x}}\right)}+1$

$\Rightarrow f^{\text{'}}\left(x\right)=\frac{1}{x}+1$

$\Rightarrow f^{\text{'}}\left(\frac{1}{x^2}\right)=x^2+1$

$\Rightarrow \sum _{x=1}^{20}{f}^{\text{'}}\left(\frac{1}{x^2}\right)=\sum _{x=1}^{20}\left(x^2+1\right)$

$\Rightarrow \sum _{x=1}^{20}{f}^{\text{'}}\left(\frac{1}{x^2}\right)=\sum _{x=1}^{20}{x}^2+20$

$\Rightarrow \sum _{x=1}^{20}{f}^{\text{'}}\left(\frac{1}{x^2}\right)=\frac{20\times 21\times 41}{6}+20$

$\Rightarrow \sum _{x=1}^{20}{f}^{\text{'}}\left(\frac{1}{x^2}\right)=2870+20$

$\Rightarrow \sum _{x=1}^{20}{f}^{\text{'}}\left(\frac{1}{x^2}\right)=2890$