លំហាត់វិសមភាព (អនុវត្តន៍វិសមភាព Schur)
Posted by van khea នៅ 02/20/2011
(van khea) ឧបមាថា 
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ដោយ
យើងតាង
យើងបានៈ
ដូចនេះយើងត្រូវស្រាយថាៈ
យើងនឹងស្រាយថាៈ
តាម វិសមភាព Schur យើងបានវិសមភាពខាងលើពិតជានិច្ច។
ដូចនេះយើងបានៈ
តែ
![\displaystyle \Rightarrow \frac{z^3+x^3-y^3}{zx}+\frac{x^3+y^3-z^3 }{xy}+\frac{y^3+z^3-x^3}{yz}\leq 2\sqrt[3]{\frac{9}{4}(a+b+c)}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5CRightarrow+%5Cfrac%7Bz%5E3%2Bx%5E3-y%5E3%7D%7Bzx%7D%2B%5Cfrac%7Bx%5E3%2By%5E3-z%5E3+%7D%7Bxy%7D%2B%5Cfrac%7By%5E3%2Bz%5E3-x%5E3%7D%7Byz%7D%5Cleq+2%5Csqrt%5B3%5D%7B%5Cfrac%7B9%7D%7B4%7D%28a%2Bb%2Bc%29%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle \Rightarrow \frac{z^3+x^3-y^3}{zx}+\frac{x^3+y^3-z^3 }{xy}+\frac{y^3+z^3-x^3}{yz}\leq 2\sqrt[3]{\frac{9}{4}(a+b+c)}")
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់ ។ សមភាពកើតមានពេល
![\displaystyle \frac{a}{\sqrt[3]{1+a^2}}+\frac{b}{\sqrt[3]{1+b^2}}+\frac{c}{\sqrt[3]{1+c^2}}\leq \sqrt[3]{\frac{9}{4}(a+b+c)}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Ba%7D%7B%5Csqrt%5B3%5D%7B1%2Ba%5E2%7D%7D%2B%5Cfrac%7Bb%7D%7B%5Csqrt%5B3%5D%7B1%2Bb%5E2%7D%7D%2B%5Cfrac%7Bc%7D%7B%5Csqrt%5B3%5D%7B1%2Bc%5E2%7D%7D%5Cleq+%5Csqrt%5B3%5D%7B%5Cfrac%7B9%7D%7B4%7D%28a%2Bb%2Bc%29%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle \frac{a}{\sqrt[3]{1+a^2}}+\frac{b}{\sqrt[3]{1+b^2}}+\frac{c}{\sqrt[3]{1+c^2}}\leq \sqrt[3]{\frac{9}{4}(a+b+c)}")
![\displaystyle \frac{a}{\sqrt[3]{ab+bc+ca+a^2}}+\frac{b}{\sqrt[3]{ab+bc+ca+b^2}}+\frac{c}{\sqrt[3]{ab+bc+ca+c^2}}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Ba%7D%7B%5Csqrt%5B3%5D%7Bab%2Bbc%2Bca%2Ba%5E2%7D%7D%2B%5Cfrac%7Bb%7D%7B%5Csqrt%5B3%5D%7Bab%2Bbc%2Bca%2Bb%5E2%7D%7D%2B%5Cfrac%7Bc%7D%7B%5Csqrt%5B3%5D%7Bab%2Bbc%2Bca%2Bc%5E2%7D%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle \frac{a}{\sqrt[3]{ab+bc+ca+a^2}}+\frac{b}{\sqrt[3]{ab+bc+ca+b^2}}+\frac{c}{\sqrt[3]{ab+bc+ca+c^2}}")
![\displaystyle \leq \sqrt[3]{\frac{9}{4}(a+b+c)}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cleq+%5Csqrt%5B3%5D%7B%5Cfrac%7B9%7D%7B4%7D%28a%2Bb%2Bc%29%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle \leq \sqrt[3]{\frac{9}{4}(a+b+c)}")
![\displaystyle \Leftrightarrow \frac{a}{\sqrt[3]{(a+b)(c+a)}}+\frac{b}{\sqrt[3]{(b+c)(a+b)}}+\frac{c}{\sqrt[3]{(c+a)(b+c)}}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5CLeftrightarrow+%5Cfrac%7Ba%7D%7B%5Csqrt%5B3%5D%7B%28a%2Bb%29%28c%2Ba%29%7D%7D%2B%5Cfrac%7Bb%7D%7B%5Csqrt%5B3%5D%7B%28b%2Bc%29%28a%2Bb%29%7D%7D%2B%5Cfrac%7Bc%7D%7B%5Csqrt%5B3%5D%7B%28c%2Ba%29%28b%2Bc%29%7D%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle \Leftrightarrow \frac{a}{\sqrt[3]{(a+b)(c+a)}}+\frac{b}{\sqrt[3]{(b+c)(a+b)}}+\frac{c}{\sqrt[3]{(c+a)(b+c)}}")
![\displaystyle \Leftrightarrow \frac{2a}{\sqrt[3]{(a+b)(c+a)}}+\frac{2b}{\sqrt[3]{(b+c)(a+b)}}+\frac{2c}{\sqrt[3]{(c+a)(b+c)}}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5CLeftrightarrow+%5Cfrac%7B2a%7D%7B%5Csqrt%5B3%5D%7B%28a%2Bb%29%28c%2Ba%29%7D%7D%2B%5Cfrac%7B2b%7D%7B%5Csqrt%5B3%5D%7B%28b%2Bc%29%28a%2Bb%29%7D%7D%2B%5Cfrac%7B2c%7D%7B%5Csqrt%5B3%5D%7B%28c%2Ba%29%28b%2Bc%29%7D%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle \Leftrightarrow \frac{2a}{\sqrt[3]{(a+b)(c+a)}}+\frac{2b}{\sqrt[3]{(b+c)(a+b)}}+\frac{2c}{\sqrt[3]{(c+a)(b+c)}}")
![\displaystyle \leq 2\sqrt[3]{\frac{9}{4}(a+b+c)}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cleq+2%5Csqrt%5B3%5D%7B%5Cfrac%7B9%7D%7B4%7D%28a%2Bb%2Bc%29%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle \leq 2\sqrt[3]{\frac{9}{4}(a+b+c)}")
![x=\sqrt[3]{a+b}; y=\sqrt[3]{b+c}; z=\sqrt[3]{c+a}](http://s0.wp.com/latex.php?latex=x%3D%5Csqrt%5B3%5D%7Ba%2Bb%7D%3B+y%3D%5Csqrt%5B3%5D%7Bb%2Bc%7D%3B+z%3D%5Csqrt%5B3%5D%7Bc%2Ba%7D&bg=fafcff&fg=2a2a2a&s=1 "x=\sqrt[3]{a+b}; y=\sqrt[3]{b+c}; z=\sqrt[3]{c+a}")
![\displaystyle \frac{z^3+x^3-y^3}{zx}+\frac{x^3+y^3-z^3 }{xy}+\frac{y^3+z^3-x^3}{yz}\leq 2\sqrt[3]{\frac{9}{4}(a+b+c)}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bz%5E3%2Bx%5E3-y%5E3%7D%7Bzx%7D%2B%5Cfrac%7Bx%5E3%2By%5E3-z%5E3+%7D%7Bxy%7D%2B%5Cfrac%7By%5E3%2Bz%5E3-x%5E3%7D%7Byz%7D%5Cleq+2%5Csqrt%5B3%5D%7B%5Cfrac%7B9%7D%7B4%7D%28a%2Bb%2Bc%29%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle \frac{z^3+x^3-y^3}{zx}+\frac{x^3+y^3-z^3 }{xy}+\frac{y^3+z^3-x^3}{yz}\leq 2\sqrt[3]{\frac{9}{4}(a+b+c)}")
![x+y+z=\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{c+a}](http://s0.wp.com/latex.php?latex=x%2By%2Bz%3D%5Csqrt%5B3%5D%7Ba%2Bb%7D%2B%5Csqrt%5B3%5D%7Bb%2Bc%7D%2B%5Csqrt%5B3%5D%7Bc%2Ba%7D&bg=fafcff&fg=2a2a2a&s=1 "x+y+z=\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{c+a}")
![\displaystyle \Leftrightarrow x+y+z\leq 3\sqrt[3]{\frac{2}{3}(a+b+c)}=2\sqrt[3]{\frac{9}{4}(a+b+c)}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5CLeftrightarrow+x%2By%2Bz%5Cleq+3%5Csqrt%5B3%5D%7B%5Cfrac%7B2%7D%7B3%7D%28a%2Bb%2Bc%29%7D%3D2%5Csqrt%5B3%5D%7B%5Cfrac%7B9%7D%7B4%7D%28a%2Bb%2Bc%29%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle \Leftrightarrow x+y+z\leq 3\sqrt[3]{\frac{2}{3}(a+b+c)}=2\sqrt[3]{\frac{9}{4}(a+b+c)}")
ជូនពរការស្រាវជ្រាវរបស់មិត្តអ្នកសិក្សាទទួលបានជោគជ័យ ពីខ្ញុំ វ៉ាន់ ឃា 
