Let for a differentiable function f:(0,) R,f(x)-f(y) _e((x)/(y))+x-y,,y (0,).

Question

Let for a differentiable function $f:(0,\infty)	o R,f(x)-f(y)\geqslant {\log }_{e}(\frac{x}{y})+x-y,\forallx,y\in (0,\infty)$. Then $\sum _{n=1}^{20}{f}^{'}(\frac{1}{{n}^{2}})$ is equal to \_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 2890

Accepted Answer

$\begin{gathered} f(x)-f(y)\ge \ln x-\ln y+x-y \ \frac{f(x)-f(y)}{x-y}\ge \frac{\ln x-\ln y}{x-y}+1 \end{gathered}$

Let $x>y$

$\underset{y\to x}{lim}{f}^{'}({x}^{-})\ge \frac{1}{x}+1\,$

Let $x$\underset{y\to x}{lim}{f}\^{'}({x}\^{+})\le \frac{1}{x}+1\,$

$$$\begin{gathered} {f}^{1}({x}^{-})={f}^{1}({x}^{+}) \ {f}^{1}(x)=\frac{1}{x}+1 \ {f}^{'}(\frac{1}{{x}^{2}})={x}^{2}+1 \end{gathered}$$$

$$$\begin{gathered} \sum _{x=1}^{20}({x}^{2}+1)=\sum _{x-1}^{20}{x}^{2}+20 \ =\frac{20\times 21\times 41}{6}+20 \ =2890 \end{gathered}$$

Solutions · 1