Problem 227 (van khea)
Posted by van khea នៅ 07/16/2011
ស្រាយបញ្ជាក់ថាចំពោះបីចំនួនពិតវិជ្ជមាន 
សំរាយបញ្ជាក់
យើងមានៈ
តាង
ដូចនេះយើងទាញបានថាៈ
តាង 
ម្យ៉ាងទៀតតាមលំហាត់ទី 32 vankhea យើងមានៈ
តាមវិសមភាព 
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។ សមភាពកើតមានពេល
![\displaystyle \frac{1}{a\sqrt[3]{pb+qc}}+\frac{1}{b\sqrt[3]{pc+qa}}+\frac{1}{c\sqrt[3]{pa+qb}}\geq \frac{3}{(abc)^{\frac{4}{9}}.\sqrt[3]{p+q}}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B1%7D%7Ba%5Csqrt%5B3%5D%7Bpb%2Bqc%7D%7D%2B%5Cfrac%7B1%7D%7Bb%5Csqrt%5B3%5D%7Bpc%2Bqa%7D%7D%2B%5Cfrac%7B1%7D%7Bc%5Csqrt%5B3%5D%7Bpa%2Bqb%7D%7D%5Cgeq+%5Cfrac%7B3%7D%7B%28abc%29%5E%7B%5Cfrac%7B4%7D%7B9%7D%7D.%5Csqrt%5B3%5D%7Bp%2Bq%7D%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle \frac{1}{a\sqrt[3]{pb+qc}}+\frac{1}{b\sqrt[3]{pc+qa}}+\frac{1}{c\sqrt[3]{pa+qb}}\geq \frac{3}{(abc)^{\frac{4}{9}}.\sqrt[3]{p+q}}")
![\displaystyle \frac{1}{a\sqrt[3]{pb+qc}}+\frac{1}{b\sqrt[3]{pc+qa}}+\frac{1}{c\sqrt[3]{pa+qb}}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B1%7D%7Ba%5Csqrt%5B3%5D%7Bpb%2Bqc%7D%7D%2B%5Cfrac%7B1%7D%7Bb%5Csqrt%5B3%5D%7Bpc%2Bqa%7D%7D%2B%5Cfrac%7B1%7D%7Bc%5Csqrt%5B3%5D%7Bpa%2Bqb%7D%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle \frac{1}{a\sqrt[3]{pb+qc}}+\frac{1}{b\sqrt[3]{pc+qa}}+\frac{1}{c\sqrt[3]{pa+qb}}")
![\displaystyle =\frac{1}{a^{\frac{2}{3}}.a^{\frac{1}{3}}\sqrt[3]{pb+qc}}+\frac{1}{b^{\frac{2}{3}}.b^{\frac{1}{3}}\sqrt[3]{pc+qa}}+\frac{1}{c^{\frac{2}{3}}.c^{\frac{1}{3}}\sqrt[3]{pa+qb}}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D%5Cfrac%7B1%7D%7Ba%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D.a%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5Csqrt%5B3%5D%7Bpb%2Bqc%7D%7D%2B%5Cfrac%7B1%7D%7Bb%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D.b%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5Csqrt%5B3%5D%7Bpc%2Bqa%7D%7D%2B%5Cfrac%7B1%7D%7Bc%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D.c%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5Csqrt%5B3%5D%7Bpa%2Bqb%7D%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle =\frac{1}{a^{\frac{2}{3}}.a^{\frac{1}{3}}\sqrt[3]{pb+qc}}+\frac{1}{b^{\frac{2}{3}}.b^{\frac{1}{3}}\sqrt[3]{pc+qa}}+\frac{1}{c^{\frac{2}{3}}.c^{\frac{1}{3}}\sqrt[3]{pa+qb}}")
![\displaystyle =\frac{1}{a^{\frac{2}{3}}\sqrt[3]{pab+qca}}+\frac{1}{b^{\frac{2}{3}}\sqrt[3]{pbc+qab}}+\frac{1}{c^{\frac{2}{3}}\sqrt[3]{pca+qbc}}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D%5Cfrac%7B1%7D%7Ba%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Csqrt%5B3%5D%7Bpab%2Bqca%7D%7D%2B%5Cfrac%7B1%7D%7Bb%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Csqrt%5B3%5D%7Bpbc%2Bqab%7D%7D%2B%5Cfrac%7B1%7D%7Bc%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Csqrt%5B3%5D%7Bpca%2Bqbc%7D%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle =\frac{1}{a^{\frac{2}{3}}\sqrt[3]{pab+qca}}+\frac{1}{b^{\frac{2}{3}}\sqrt[3]{pbc+qab}}+\frac{1}{c^{\frac{2}{3}}\sqrt[3]{pca+qbc}}")
![\displaystyle =\frac{(bc)^{\frac{2}{3}}}{k^{\frac{2}{3}}\sqrt[3]{pab+qca}}+\frac{(ca)^{\frac{2}{3}}}{k^{\frac{2}{3}}\sqrt[3]{pbc+qab}}+\frac{(ab)^{\frac{2}{3}}}{k^{\frac{2}{3}}\sqrt[3]{pca+qbc}}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D%5Cfrac%7B%28bc%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%7D%7Bk%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Csqrt%5B3%5D%7Bpab%2Bqca%7D%7D%2B%5Cfrac%7B%28ca%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%7D%7Bk%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Csqrt%5B3%5D%7Bpbc%2Bqab%7D%7D%2B%5Cfrac%7B%28ab%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%7D%7Bk%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Csqrt%5B3%5D%7Bpca%2Bqbc%7D%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle =\frac{(bc)^{\frac{2}{3}}}{k^{\frac{2}{3}}\sqrt[3]{pab+qca}}+\frac{(ca)^{\frac{2}{3}}}{k^{\frac{2}{3}}\sqrt[3]{pbc+qab}}+\frac{(ab)^{\frac{2}{3}}}{k^{\frac{2}{3}}\sqrt[3]{pca+qbc}}")
![\displaystyle =\frac{1}{k^{\frac{2}{3}}}\biggl(\frac{x^2}{\sqrt[3]{pz^3+qy^3}}+\frac{y^2}{\sqrt[3]{px^3+qz^3}}+\frac{z^2}{\sqrt[3]{py^3+qx^3}}\biggl)](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D%5Cfrac%7B1%7D%7Bk%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%7D%5Cbiggl%28%5Cfrac%7Bx%5E2%7D%7B%5Csqrt%5B3%5D%7Bpz%5E3%2Bqy%5E3%7D%7D%2B%5Cfrac%7By%5E2%7D%7B%5Csqrt%5B3%5D%7Bpx%5E3%2Bqz%5E3%7D%7D%2B%5Cfrac%7Bz%5E2%7D%7B%5Csqrt%5B3%5D%7Bpy%5E3%2Bqx%5E3%7D%7D%5Cbiggl%29&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle =\frac{1}{k^{\frac{2}{3}}}\biggl(\frac{x^2}{\sqrt[3]{pz^3+qy^3}}+\frac{y^2}{\sqrt[3]{px^3+qz^3}}+\frac{z^2}{\sqrt[3]{py^3+qx^3}}\biggl)")
![\displaystyle =\frac{1}{k^{\frac{2}{3}}}\biggl(\frac{(x^2)^{\frac{4}{3}}}{\sqrt[3]{pz^3x^2+qx^2y^3}}+\frac{(y^2)^{\frac{4}{3}}}{\sqrt[3]{px^3y^2+qy^2z^3}}+\frac{(z^2)^{\frac{4}{3}}}{\sqrt[3]{py^3z^2+qz^2x^3}}\biggl)](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D%5Cfrac%7B1%7D%7Bk%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%7D%5Cbiggl%28%5Cfrac%7B%28x%5E2%29%5E%7B%5Cfrac%7B4%7D%7B3%7D%7D%7D%7B%5Csqrt%5B3%5D%7Bpz%5E3x%5E2%2Bqx%5E2y%5E3%7D%7D%2B%5Cfrac%7B%28y%5E2%29%5E%7B%5Cfrac%7B4%7D%7B3%7D%7D%7D%7B%5Csqrt%5B3%5D%7Bpx%5E3y%5E2%2Bqy%5E2z%5E3%7D%7D%2B%5Cfrac%7B%28z%5E2%29%5E%7B%5Cfrac%7B4%7D%7B3%7D%7D%7D%7B%5Csqrt%5B3%5D%7Bpy%5E3z%5E2%2Bqz%5E2x%5E3%7D%7D%5Cbiggl%29&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle =\frac{1}{k^{\frac{2}{3}}}\biggl(\frac{(x^2)^{\frac{4}{3}}}{\sqrt[3]{pz^3x^2+qx^2y^3}}+\frac{(y^2)^{\frac{4}{3}}}{\sqrt[3]{px^3y^2+qy^2z^3}}+\frac{(z^2)^{\frac{4}{3}}}{\sqrt[3]{py^3z^2+qz^2x^3}}\biggl)")
![\displaystyle \geq \frac{(x^2+y^2+z^2)^{\frac{4}{3}}}{k^{\frac{2}{3}}.\sqrt[3]{p(x^3y^2+y^3z^2+z^3x^2)+q(x^2y^3+y^2z^3+z^2x^3)}}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cgeq+%5Cfrac%7B%28x%5E2%2By%5E2%2Bz%5E2%29%5E%7B%5Cfrac%7B4%7D%7B3%7D%7D%7D%7Bk%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D.%5Csqrt%5B3%5D%7Bp%28x%5E3y%5E2%2By%5E3z%5E2%2Bz%5E3x%5E2%29%2Bq%28x%5E2y%5E3%2By%5E2z%5E3%2Bz%5E2x%5E3%29%7D%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle \geq \frac{(x^2+y^2+z^2)^{\frac{4}{3}}}{k^{\frac{2}{3}}.\sqrt[3]{p(x^3y^2+y^3z^2+z^3x^2)+q(x^2y^3+y^2z^3+z^2x^3)}}")
![\displaystyle \Rightarrow \frac{1}{a\sqrt[3]{pb+qc}}+\frac{1}{b\sqrt[3]{pc+qa}}+\frac{1}{c\sqrt[3]{pa+qb}}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5CRightarrow+%5Cfrac%7B1%7D%7Ba%5Csqrt%5B3%5D%7Bpb%2Bqc%7D%7D%2B%5Cfrac%7B1%7D%7Bb%5Csqrt%5B3%5D%7Bpc%2Bqa%7D%7D%2B%5Cfrac%7B1%7D%7Bc%5Csqrt%5B3%5D%7Bpa%2Bqb%7D%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle \Rightarrow \frac{1}{a\sqrt[3]{pb+qc}}+\frac{1}{b\sqrt[3]{pc+qa}}+\frac{1}{c\sqrt[3]{pa+qb}}")
![\displaystyle \geq \frac{\sqrt{3}(x^2+y^2+z^2)^{\frac{4}{3}}}{k^{\frac{2}{3}}\sqrt[3]{p+q}(x^2+y^2+z^2)^{\frac{5}{6}}}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cgeq+%5Cfrac%7B%5Csqrt%7B3%7D%28x%5E2%2By%5E2%2Bz%5E2%29%5E%7B%5Cfrac%7B4%7D%7B3%7D%7D%7D%7Bk%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Csqrt%5B3%5D%7Bp%2Bq%7D%28x%5E2%2By%5E2%2Bz%5E2%29%5E%7B%5Cfrac%7B5%7D%7B6%7D%7D%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle \geq \frac{\sqrt{3}(x^2+y^2+z^2)^{\frac{4}{3}}}{k^{\frac{2}{3}}\sqrt[3]{p+q}(x^2+y^2+z^2)^{\frac{5}{6}}}")
![\displaystyle \geq \frac{\sqrt{3}(x^2+y^2+z^2)^{\frac{1}{2}}}{k^{\frac{2}{3}}\sqrt[3]{p+q}}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cgeq+%5Cfrac%7B%5Csqrt%7B3%7D%28x%5E2%2By%5E2%2Bz%5E2%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7Bk%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Csqrt%5B3%5D%7Bp%2Bq%7D%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle \geq \frac{\sqrt{3}(x^2+y^2+z^2)^{\frac{1}{2}}}{k^{\frac{2}{3}}\sqrt[3]{p+q}}")
![\displaystyle x^2+y^2+z^2\geq 3\sqrt[3]{x^2y^2z^2}=3\sqrt[3]{(abc)^{\frac{4}{3}}}=3k^{\frac{4}{9}}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+x%5E2%2By%5E2%2Bz%5E2%5Cgeq+3%5Csqrt%5B3%5D%7Bx%5E2y%5E2z%5E2%7D%3D3%5Csqrt%5B3%5D%7B%28abc%29%5E%7B%5Cfrac%7B4%7D%7B3%7D%7D%7D%3D3k%5E%7B%5Cfrac%7B4%7D%7B9%7D%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle x^2+y^2+z^2\geq 3\sqrt[3]{x^2y^2z^2}=3\sqrt[3]{(abc)^{\frac{4}{3}}}=3k^{\frac{4}{9}}")
![\displaystyle \Rightarrow \frac{1}{a\sqrt[3]{pb+qc}}+\frac{1}{b\sqrt[3]{pc+qa}}+\frac{1}{c\sqrt[3]{pa+qb}}\geq \frac{3k^{\frac{2}{9}}}{k^{\frac{2}{3}}\sqrt[3]{p+q}}=\frac{3}{(abc)^{\frac{4}{9}}.\sqrt[3]{p+q}}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5CRightarrow+%5Cfrac%7B1%7D%7Ba%5Csqrt%5B3%5D%7Bpb%2Bqc%7D%7D%2B%5Cfrac%7B1%7D%7Bb%5Csqrt%5B3%5D%7Bpc%2Bqa%7D%7D%2B%5Cfrac%7B1%7D%7Bc%5Csqrt%5B3%5D%7Bpa%2Bqb%7D%7D%5Cgeq+%5Cfrac%7B3k%5E%7B%5Cfrac%7B2%7D%7B9%7D%7D%7D%7Bk%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Csqrt%5B3%5D%7Bp%2Bq%7D%7D%3D%5Cfrac%7B3%7D%7B%28abc%29%5E%7B%5Cfrac%7B4%7D%7B9%7D%7D.%5Csqrt%5B3%5D%7Bp%2Bq%7D%7D&bg=fafcff&fg=2a2a2a&s=1 "\displaystyle \Rightarrow \frac{1}{a\sqrt[3]{pb+qc}}+\frac{1}{b\sqrt[3]{pc+qa}}+\frac{1}{c\sqrt[3]{pa+qb}}\geq \frac{3k^{\frac{2}{9}}}{k^{\frac{2}{3}}\sqrt[3]{p+q}}=\frac{3}{(abc)^{\frac{4}{9}}.\sqrt[3]{p+q}}")
Long បាននិយាយ
សួស្តីបង! សុខសប្បាយទេ?
vankhea បាននិយាយ
ដូចជាបាត់មុខបាត់មាត់យូរនែ!
ខ្ញុំសុខសប្បាយជាធម្មតាទេ ចំនែកគាឡុងវិញសុខទុកយ៉ាងណាដែរ ?