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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

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

1 Answer 1

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This might look very silly, but:

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

This gives me the idea that: $f(x) = \ln x + x + C$ (where $C$ is some random constant).
The rest is obvious.

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