# Value of $\sum\limits^{20}_{n=1}f'\bigg(\frac{1}{n^2}\bigg)=$

Let for a differentiable function $$\displaystyle f:\bigg(0,\infty\bigg)\rightarrow \mathbb{R}$$ and $$\displaystyle f(x)-f(y)\geq \ln\bigg(\frac{x}{y}\bigg)+x-y$$ for all $$x\in(0,\infty)$$. Then $$\displaystyle \sum^{20}_{n=1}f'\bigg(\frac{1}{n^2}\bigg)$$

What I try :

$$\displaystyle f(x)-f(y)\geq \ln\bigg(\frac{x}{y}\bigg)+x-y\cdots (1)$$

Now interchange $$x\rightarrow y$$, Then

$$\displaystyle f(y)-f(x)\leq \ln\bigg(\frac{y}{x}\bigg)+y-x$$

$$\displaystyle f(x)-f(y)\leq \ln\bigg(\frac{x}{y}\bigg)+x-y\cdots (2)$$

Form $$(1)$$ and $$(2)$$, We get

$$\displaystyle f(x)-f(y)= \ln\bigg(\frac{x}{y}\bigg)+x-y$$

How do I solve it , please have a look, Thanks

• Differentiating the last equality gives $f'(x) = 1/x + 1$. Feb 12 at 13:19
• Does your inequality $f(x) - f(y) \ge \cdots$ valid for some $y$ or all $y$? In the second case, the inequality will force $f(x) = \log(x) + x + const.$. Feb 12 at 15:51

$$f(x)-f(y) = \ln{(\frac{x}{y})} + x - y = \ln{x} - \ln{y} + x - y$$
This gives me the idea that: $$f(x) = \ln x + x + C$$ (where $$C$$ is some random constant).