Problem 262 (vankhea)
Posted by van khea នៅ 08/02/2011
ឧបមាថាគេមានបីចំនួនពិតវិជ្ជមាន 
សំរាយបញ្ជាក់
វិសមភាពខាងលើសមមូលនឹងៈ
ម្យ៉ាងទៀតយើងមានៈ
ដូចនេះយើងត្រូវស្រាយថាៈ
តាមវិសមភាព 
គុណអង្គនិងអង្គនៃវិសមភាពខាងលើយើងបានៈ
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។ សមភាពកើតមានពេល
![3+x+y+z=1+x+1+y+1+z\geq 3\sqrt[3]{(1+x)(1+y)(1+z)}](http://s0.wp.com/latex.php?latex=3%2Bx%2By%2Bz%3D1%2Bx%2B1%2By%2B1%2Bz%5Cgeq+3%5Csqrt%5B3%5D%7B%281%2Bx%29%281%2By%29%281%2Bz%29%7D&bg=fafcff&fg=2a2a2a&s=1 "3+x+y+z=1+x+1+y+1+z\geq 3\sqrt[3]{(1+x)(1+y)(1+z)}")
![(1+x)(1+y)+(1+y)(1+z)+(1+z)(1+x)\geq 3\sqrt[3]{(1+x)^2(1+y)^2(1+z)^2}](http://s0.wp.com/latex.php?latex=%281%2Bx%29%281%2By%29%2B%281%2By%29%281%2Bz%29%2B%281%2Bz%29%281%2Bx%29%5Cgeq+3%5Csqrt%5B3%5D%7B%281%2Bx%29%5E2%281%2By%29%5E2%281%2Bz%29%5E2%7D&bg=fafcff&fg=2a2a2a&s=1 "(1+x)(1+y)+(1+y)(1+z)+(1+z)(1+x)\geq 3\sqrt[3]{(1+x)^2(1+y)^2(1+z)^2}")
