H a p p y 1 9 !
$$\frac{\it{3}}{\it{3}+ \it{2}}+ \frac{\it{3}^{\,\it{2}}}{\it{3}^{\,\it{2}}+ \it{2}}+ \,...\,+ \frac{\it{3}^{\,\it{2019}}}{\it{3}^{\,\it{2019}}+ \it{2}}> \frac{\it{2019}^{\,\it{2019}- \it{2018}}}{\it{2020}}$$
H a p p y 1 9 !
$$\frac{\it{3}}{\it{3}+ \it{2}}+ \frac{\it{3}^{\,\it{2}}}{\it{3}^{\,\it{2}}+ \it{2}}+ \,...\,+ \frac{\it{3}^{\,\it{2019}}}{\it{3}^{\,\it{2019}}+ \it{2}}> \frac{\it{2019}^{\,\it{2019}- \it{2018}}}{\it{2020}}$$
H a p p y 1 9 !
$$\frac{\it{3}}{\it{3}+ \it{2}}+ \frac{\it{3}^{\,\it{2}}}{\it{3}^{\,\it{2}}+ \it{2}}+ \,...\,+ \frac{\it{3}^{\,\it{2019}}}{\it{3}^{\,\it{2019}}+ \it{2}}> \frac{\it{2019}^{\,\it{2019}- \it{2018}}}{\it{2020}}$$
Vế trái $> \frac{3}{3+2}+\frac{3^2}{3^2+2}> \frac{1}{2}+\frac{1}{2}=1> \frac{2019}{2020}=\frac{2019^{2019-2018}}{2020}=$ Vế phải
...
Ðêm nay tiễn đưa
Giây phút cuối vẫn còn tay ấm tay
Mai sẽ thấm cơn lạnh khi gió lay
Và những lúc mưa gọi thương nhớ đầy ...
H a p p y 1 9 !
$$\frac{\it{3}}{\it{3}+ \it{2}}+ \frac{\it{3}^{\,\it{2}}}{\it{3}^{\,\it{2}}+ \it{2}}+ \,...\,+ \frac{\it{3}^{\,\it{2019}}}{\it{3}^{\,\it{2019}}+ \it{2}}> \frac{\it{2019}^{\,\it{2019}- \it{2018}}}{\it{2020}}$$
H a p p y 1 9 !
$$\frac{\it{3}}{\it{3}+ \it{2}}+ \frac{\it{3}^{\,\it{2}}}{\it{3}^{\,\it{2}}+ \it{2}}+ \,...\,+ \frac{\it{3}^{\,\it{2019}}}{\it{3}^{\,\it{2019}}+ \it{2}}> \frac{\it{2019}^{\,\it{2020}- \it{2018}}}{\it{2020}}$$
H a p p y 1 9 !
$$\frac{\it{3}}{\it{3}+ \it{2}}+ \frac{\it{3}^{\,\it{2}}}{\it{3}^{\,\it{2}}+ \it{2}}+ \,...\,+ \frac{\it{3}^{\,\it{2019}}}{\it{3}^{\,\it{2019}}+ \it{2}}> \frac{\it{2019}^{\,\it{2020}- \it{2018}}}{\it{2020}}$$
Điều cần chứng minh tương đương với :
$2019-\left ( \frac{3}{3+2}+\frac{3^2}{3^2+2}+...+\frac{3^{2019}}{3^{2019}+2} \right )< 2019-\frac{2019^2}{2020}$
$\Leftrightarrow \frac{2}{3+2}+\frac{2}{3^2+2}+\frac{2}{3^3+2}+...+\frac{2}{3^{2019}+2}< \frac{2019}{2020}$ $(^\ast )$
Vậy ta cần chứng minh $(^\ast )$.
Ta có :
$\frac{2}{3+2}+\frac{2}{3^2+2}+\frac{2}{3^3+2}+...+\frac{2}{3^{2019}+2}< \frac{2}{5}+\left ( \frac{2}{3^2}+\frac{2}{3^3}+...+\frac{2}{3^{2019}} \right )=\frac{2}{5}+\frac{1}{3}\left [ 1-\left ( \frac{1}{3} \right )^{2018} \right ]< \frac{2}{5}+\frac{1}{3}< \frac{3}{6}+\frac{1}{3}=\frac{5}{6}< \frac{2019}{2020}$
Bài toán đã được chứng minh.
...
Ðêm nay tiễn đưa
Giây phút cuối vẫn còn tay ấm tay
Mai sẽ thấm cơn lạnh khi gió lay
Và những lúc mưa gọi thương nhớ đầy ...
H a p p y 1 9 !
$$\frac{\it{3}}{\it{3}+ \it{2}}+ \frac{\it{3}^{\,\it{2}}}{\it{3}^{\,\it{2}}+ \it{2}}+ \,...\,+ \frac{\it{3}^{\,\it{2019}}}{\it{3}^{\,\it{2019}}+ \it{2}}> \frac{\it{2019}^{\,\it{2020}- \it{2018}}}{\it{2020}}$$
Trước hết , ta luôn có với mọi số tự nhiên $\it{k}> \it{2}$ , thì :
$\frac{\it{1}}{\it{3}^{\,\it{k}}}- \frac{\it{54}}{\it{29}\,.\,\it{3}^{\,\it{2}\,\it{k}}}- \frac{\it{1}}{\it{3}^{\,\it{k}}+ \it{2}}= \frac{\it{4}\left ( \it{3}^{\,\it{k}}- \it{27} \right )}{\it{29}\,.\,\it{3}^{\,\it{2}\,\it{k}}\left ( \it{3}^{\,\it{k}}+ \it{2} \right )}\geqq 0$
$\frac{\it{1}}{\it{3}+ \it{2}}+ \frac{\it{1}}{\it{3}^{\,\it{2}}+ \it{2}}+ \,...\,+ \frac{\it{1}}{\it{3}^{\,\it{2019}}+ \it{2}}< \frac{\it{1}}{\it{3}+ \it{2}}+ \frac{\it{1}}{\it{3}^{\,\it{2}}+ \it{2}}+ \sum\limits_{\it{k}= \it{3}}^{\it{2019}} \left ( \frac{\it{1}}{\it{3}^{\,\it{k}}}- \frac{\it{54}}{\it{29}\,.\,\it{3}^{\,\it{2}\,\it{k}}} \right )< \frac{\it{1}}{\it{3}+ \it{2}}+ \frac{\it{1}}{\it{3}^{\,\it{2}}+ \it{2}}+ \it{0}\,.\,\it{0526819992337165}< \it{0}\,.\,\it{34359108\,...\,703\left ( 90 \right )}$
$\Leftrightarrow \frac{\it{3}}{\it{3}+ \it{2}}+ \frac{\it{3}^{\,\it{2}}}{\it{3}^{\,\it{2}}+ \it{2}}+ \,...\,+ \frac{\it{3}^{\,\it{2019}}}{\it{3}^{\,\it{2019}}+ \it{2}}> \frac{\it{2019}^{\,\it{2020}- \it{2018}}}{\it{2020}}$
* N h ậ n x é t :
$\sum\limits_{\it{k}= \it{1}}^{+ \infty }\,\left ( \frac{\it{1}}{\it{3}^{\,\it{k}}}- \frac{\it{54}}{\it{29}\,.\,\it{3}^{\,\it{2}\,\it{k}}} \right )= \underbrace{\frac{\it{1}}{\it{116}}\,.\,\it{3}^{-\,\it{2}\,\it{n}}\left ( -\,\it{58}\,.\,\it{3}\,n+ \it{31}\,.\,\it{3}^{\,\it{2}\,\it{n}}+ \it{27} \right )}_{\it{n}\rightarrow + \infty }= \frac{\it{31}}{\it{116}}$
$\sum\limits_{\it{k}= \it{1}}^{+ \infty }\,\frac{\it{1}}{\it{3}^{\,\it{k}}+ \it{2}}= \it{0}\,.\,\it{343575065043899}\,...$
0 thành viên, 2 khách, 0 thành viên ẩn danh