វិសមភាពក្នុងការប្រឡង PSFJIMO-USA 1993

ប្រៀបធៀបពីរចំនួន

លំហាត់វិសមភាព ក្នុងការប្រឡង CMO 1998

លំហាត់ប្រចាំថ្ងៃ៖ វិសមភាពក្នុងធរណីមាត្រ

46th Canada MO: Problem Number 1

ប្រធានលំហាត់៖

គេឲ្យ $ a_1,\ a_2,\ ...,\ a_n$ ជាបណ្តាចំនួនពិតវិជ្ជមាន ដែលមានផលគុណស្មើ 1។ 

បង្ហាញថាផលបូកៈ

\[\dfrac{a_1}{1+a_1}+\dfrac{a_2}{(1+a_1)(1+a_2)}+\dfrac{a_3}{(1+a_1)(1+a_2)(1+a_3)}+...+\dfrac{a_n}{(1+a_1)(1+a_2)...(1+a_n)}\]

គឺធំជាង រឺ ស្មើនឹង $\dfrac{2^n-1}{2^n}$ ។

ដំណោះស្រាយ

$​​​\dfrac{a_m}{(1+a_1)(1+a_2)...(1+a_m)}$

$=\dfrac{1+a_m}{(1+a_1)(1+a_2)...(1+a_m)}-\dfrac{1}{(1+a_1)(1+a_2)...(1+a_m)}$

$=\dfrac{1}{(1+a_1)...(1+a_{m-1})}-\dfrac{1}{(1+a_1)...(1+a_m)}$

ដូចនេះ បើយើងយក $b_j=(1+a_1)(1+a_2)...(1+a_j)$, ដែល $b_0=0$, ពេលនោះយើងបានផលបូក

\[\sum_{j=1}^{n}\dfrac{a_j}{(1+a_1)...(1+a_j)}=\sum_{j=1}^{n}(\dfrac{1}{b_j-1}-\dfrac{1}{b_j})=1-\dfrac{1}{b_n}\]

យកចិត្តទុកដាក់ថា $b_n=(1+a_1)(1+a_2)...(1+a_n)\ge(2\sqrt{a_1})(2\sqrt{a_2})...(2\sqrt{a_n})=2^n$

ដែល សមភាពកើតមានលុះត្រាតែគ្រប់ $a_i$ ស្មើនឹង $1$ ។ ដូចនេះ,

\[1-\dfrac{1}{b_n}\ge 1-\dfrac{1}{2^n}=\dfrac{2^n-1}{2^n}\]

ដើម្បីដឹងថាតម្លៃតូចបំផុតនេះ អាចកើតមានរឺទេ, គឺយើងជំនួនគ្រប់ $a_i=1$ ចូល យើងបានៈ

\[\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^n}=\dfrac{2^{n-1}+2^{n-2}+...+1}{2^n}=\dfrac{2^n-1}{2^n}\]

បញ្ហាត្រូវបានស្រាយបញ្ជាក់។

Maths Exercise #42: Inequality fro Balkan MO 2014

ប្រធានលំហាត់៖
គេឲ្យ $x,\ y$ និង $z$ ជាចំនួនពិតវិជ្ជមានបី ផ្ទៀងផ្ទាត់ $xy+yz+zx=3xyz$។
បង្ហាញថាៈ  $x^2y+y^2z+z^2x\ge 2(x+y+z)-3$ ហើយសមភាពកើតមាននៅពេលណា?

ដំណោះស្រាយ
លក្ខខណ្ឌដែលឲ្យអាចសរសេរជា $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3$
តាមនេះ, យើងបានៈ
\[ x^2y+y^2z+z^2x-2(x+y+z)+3=x^2y-2x+\frac{1}{y}+y^2z-2y+\frac{1}{z}+z^2x-2x+\frac{1}{x}\\=y\left(x-\frac{1}{y}\right)^2+z\left(y-\frac{1}{z}\right)^2+x\left(z-\frac{1}{z}\right)^2\ge 0\]
វិសមភាពកើតមានលុះត្រាតែ $xy=yz=zx=1$ រឺនៅពេល $x=y=z=1$

Maths Exercise #41: Requested translate exercise, Inequality

ប្រធានលំហាត់៖
គេតាង $a,b,c$ ជាប្រវែងជ្រុងទាំងបី ហើយ $x,y,z$ ជាប្រវែងបន្ទាត់ពុះក្នុងទាំងបីរបស់ត្រីកោណ $\Delta ABC$។
ចូរស្រាយបញ្ជាក់ថាៈ $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$

ដំណោះស្រាយ
តាង $AI=x$ ជាបន្ទាត់ពុះក្នុងរបស់ $\Delta ABC$ ,យើងបានក្រឡាផ្ទៃ$\Delta ABC$ ស្មើនឹងៈ
\[S=\dfrac{1}{2}bc\sin A=\dfrac{1}{2}xc\sin\dfrac{A}{2}+\dfrac{1}{2}xb\sin\dfrac{A}{2}\\ \Rightarrow 2bc\sin\dfrac{A}{2}\cos\dfrac{A}{2}=x(b+c)\sin\dfrac{A}{2}\\ \Rightarrow x=\dfrac{2bc\cos\dfrac{A}{2}}{b+c}<\dfrac{2bc}{b+c}\ \Rightarrow\ \dfrac{1}{x}>\dfrac{b+c}{2bc}\ \Rightarrow\ \dfrac{1}{x}>\dfrac{1}{2}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\quad (1)\]
ដូចគ្នាដែរ យើងបានៈ $\dfrac{1}{y}>\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{c}\right)\ \ \  (2);\quad \dfrac{1}{z}>\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ \ \ (3)$
បូកអង្គនឹងអង្គនៃវិសមភាព$(1),(2),(3)$ យើងបានៈ
\[\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\]

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៖
https://www.facebook.com/photo.php?fbid=775472742477186&set=gm.244297135774210&type=1

Maths Exercise #38: Two requested exercises

Inequality, request by Vatana Kitty Love

Limit, request by Ga Ara

Free Maths Book: Inequality collection in Maths Forum (VN language)

For who like to use VN maths book to study and prepare for outstanding student test it good to use this book to study inequality. It a good book that have nearly all kind of inequality in high school maths, was collected by a group of VN teacher and VN outstanding student in their maths forum.

Hope u can use it as a useful documents in your studying.

About this book:

+Tittle:  Tuyen Tap Bat Dang Thuc

+Size: 1.84MB

+File Type: PDF

+Link Download: Dropbox File

Maths Exercise #26: Answer for VNMO 2014-Day 2-P2

Maths Exercise #25: Good one from CMO, 2000

Solution by brother Somaisatchya Un

Maths Exercise #16: New inequality for 2014 test ^_^

In 2014 test, hope this test will be selected. So now let try your skill to solve it.

Here is the full solution from brother, Chakrey Duong, let see his solving skill ^_^

Maths Exercise #10: Geometry Again ^_^

Enjoy inequality in geometry, this exercise i choose from geometry book for grade 8 in VN book.