Let for a differentiable function f: (0,x/y) + x − y, ∀ x ∞) → R, f(x) − f(y) ≥ loge(, y ∈ (0, ∞)

Question

Let for a differentiable function \(\mathrm{f}:(0, \infty) \rightarrow \mathrm{R}\), \(\mathrm {f(x)-f(y) \geq \log _{e}\left(\frac{x}{y}\right)+x-y, \forall x, y \in(0, }\infty)\). Then \(\sum\limits_{\mathrm{n}=1}^{20} \mathrm{f}^{\prime}\left(\frac{1}{\mathrm{n}^{2}}\right)\) is equal to _____.

Answer 1

Correct answer: 2890

\(\mathrm{f}(\mathrm{x})-\mathrm{f}(\mathrm{y}) \geq \ln \mathrm{x}-\ln \mathrm{y}+\mathrm{x}-\mathrm{y}\)

\(\mathrm{\frac{f(x)-f(y)}{x-y} \geq \frac{\ln x-\ln y}{x-y}+1}\)

Let \(\mathrm{x}>\mathrm{y}\)

\(\lim\limits _{y \rightarrow x} f^{\prime}\left(x^{-}\right) \geq \frac{1}{x}+1\)

Let \(\mathrm{x}<\mathrm{y}\)

\(\mathrm {\lim\limits_{y \rightarrow x} f^{\prime}\left(x^{+}\right) \leq \frac{1}{x}+1}\)

\(\mathrm{f}^{1}\left(\mathrm{x}^{-}\right)=\mathrm{f}^{1}\left(\mathrm{x}^{+}\right)\)

\(\mathrm{f}^{1}(\mathrm{x})=\frac{1}{\mathrm{x}}+1\)

\(f^{\prime}\left(\frac{1}{x^{2}}\right)=x^{2}+1\)

\(\sum\limits_{\mathrm{x}=1}^{20}\left(\mathrm{x}^{2}+1\right)=\sum\limits_{\mathrm{x}-1}^{20} \mathrm{x}^{2}+20\)

\(=\frac{20 \times 21 \times 41}{6}+20\)

\(=2890\)