Let for a differentiable function $\mathrm{f}:(0,\mathrm{\infty <!--\; \infty \; -->})\to <!--\; \to \; -->\mathrm{R}$, $\mathrm{f}(\mathrm{x})-<!--\; -\; -->\mathrm{f}(\mathrm{y})\ge <!--\; \ge \; -->{\log }_{\mathrm{e}}<!--\; \; -->\left(\frac{\mathrm{x}}{\mathrm{y}}\right)+\mathrm{x}-<!--\; -\; -->\mathrm{y},\mathrm{\forall <!--\; \forall \; -->}\mathrm{x},\mathrm{y}\in <!--\; \in \; -->(0,\mathrm{\infty <!--\; \infty \; -->})$. Then $\underset{\mathrm{n}=1}{\overset{20}{\sum <!--\; \sum \; -->}}{\mathrm{f}}^{\mathrm{\prime <!--\; \prime \; -->}}\left(\frac{1}{{\mathrm{n}}^2}\right)$ is equal to ____

The correct answer is 2890.

We are given a differentiable function $f:(0,\infty )\to R$ satisfying:

$f\left(x\right)-f\left(y\right)\ge {\log }_e\left(\frac{x}{y}\right)+x-y,\forall x,y\in (0,\infty )$

Step 1: Derive a bound from the inequality by swapping x and y.

The given condition holds for all x, y in (0, ∞). If we swap x and y, we get:

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

Multiplying both sides by -1 (and flipping the inequality):

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

Step 2: Combine both inequalities to get an equality.

From the original condition we have $f\left(x\right)-f\left(y\right)\ge {\log }_e\left(\frac{x}{y}\right)+x-y$, and from Step 1 we have $f\left(x\right)-f\left(y\right)\le {\log }_e\left(\frac{x}{y}\right)+x-y$. Therefore:

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

Step 3: Differentiate with respect to x to find f'(x).

Differentiating both sides with respect to x (treating y as a constant):

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

Step 4: Compute f'(1/n²).

Substituting $x=\frac{1}{n^2}$:

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

Step 5: Evaluate the required summation.

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

Using the standard formula $\sum _{n=1}^N{n}^2=\frac{N(N+1)(2N+1)}{6}$ with N = 20:

$\sum _{n=1}^{20}{n}^2=\frac{20\times 21\times 41}{6}=\frac{17220}{6}=2870$

And:

$\sum _{n=1}^{20}1=20$

Therefore:

$\sum _{n=1}^{20}f\text{'}\left(\frac{1}{n^2}\right)=2870+20=$2890