Maths Exercise #40: Inequality in 2010 version ^_^

ប្រធានលំហាត់៖

ដំណោះស្រាយ
យើងមានៈ
$\frac{1}{x_1+2010}+\frac{1}{x_2+2010}+\frac{1}{x_3+2010}+\frac{1}{x_4+2010}+\frac{1}{x_5+2010}=\frac{1}{2010}\\ \Leftrightarrow \frac{1}{x_2+2010}+\frac{1}{x_3+2010}+\frac{1}{x_4+2010}+\frac{1}{x_5+2010}=\frac{1}{2010}-\frac{1}{x_1+2010}=\frac{x_1}{(x_1+2010).2010}$

អនុវត្តន៍វិសមភាពកូស៊ី, យើងបានៈ
$\frac{1}{x_2+2010}+\frac{1}{x_3+2010}+\frac{1}{x_4+2010}+\frac{1}{x_5+2010}\ge\\ 4\sqrt[4]{\frac{1}{x_2+2010}.\frac{1}{x_3+2010}.\frac{1}{x_4+2010}.\frac{1}{x_5+2010}}$

ពេលនោះ យើងបានៈ
$\frac{x_1}{(x_1+2010).2010}\ge 4.\sqrt[4]{\frac{1}{x_2+2010}.\frac{1}{x_3+2010}.\frac{1}{x_4+2010}.\frac{1}{x_5+2010}}\quad (1)$

ធ្វើដូចគ្នាដែរ យើងបានៈ
$\frac{x_2}{(x_2+2010).2010}\ge 4.\sqrt[4]{\frac{1}{x_1+2010}.\frac{1}{x_3+2010}.\frac{1}{x_4+2010}.\frac{1}{x_5+2010}}\quad (2)\\ \frac{x_3}{(x_3+2010).2010}\ge 4.\sqrt[4]{\frac{1}{x_1+2010}.\frac{1}{x_2+2010}.\frac{1}{x_4+2010}.\frac{1}{x_5+2010}}\quad (3)\\ \frac{x_4}{(x_4+2010).2010}\ge 4.\sqrt[4]{\frac{1}{x_1+2010}.\frac{1}{x_2+2010}.\frac{1}{x_3+2010}.\frac{1}{x_5+2010}}\quad (4)\\ \frac{x_5}{(x_5+2010).2010}\ge 4.\sqrt[4]{\frac{1}{x_1+2010}.\frac{1}{x_2+2010}.\frac{1}{x_3+2010}.\frac{1}{x_4+2010}}\quad (5)$

យក $(1)\times(2)\times(3)\times(4)\times(5)$ អង្គនឹងអង្គ យើងបានៈ
$\frac{x_1.x_2.x_3.x_4.x_5}{(x_1+2010)(x_2+2010)(x_3+2010)(x_4+2010)(x_5+2010).2010^5}\\ \ge4^5.\sqrt[4]{\frac{1}{(x_1+2010)^4}.\frac{1}{(x_2+2010)^4}.\frac{1}{(x_3+2010)^4}.\frac{1}{(x_4+2010)^4}.\frac{1}{(x_5+2010)^4}}\\ \Leftrightarrow x_1.x_2.x_3.x_4.x_5\ge 2010^5.4^5$

ដូចនេះ  $\sqrt[5]{x_1.x_2.x_3.x_4.x_5}\ge 8040$
សញ្ញា $"="$ កើតមានពេលៈ $x_1=x_2=x_3=x_4=x_5=8040$

មើលដំណោះស្រាយរបស់ប្អូន Punrong Rany តាមតំនភ្ជាប់ខាងក្រោម៖
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