# Problem 258 (vankhea)

ស្រាយបញ្ជាក់ថាចំពោះបណ្ដាចំនួនពិតវិជ្ជមាន $a, b, c$ គេបានៈ
$\displaystyle \frac{7}{3a}+\frac{7}{3b}+\frac{7}{3c}+\frac{15}{a+b+c}$$\displaystyle \geq \frac{6}{2a+b}+\frac{6}{2b+c}+\frac{6}{2c+a}+\frac{6}{a+2b}+\frac{6}{b+2c}+\frac{6}{c+2a}$
សំរាយបញ្ជាក់
អនុវត្តន៍ Problem 255 van khea យើងមានៈ
$\displaystyle x^3+y^3+z^3+3xyz\geq \frac{2}{9}(x+y+z)^3$
$\displaystyle \Leftrightarrow 7(x^3+y^3+z^3)+15xyz\geq 6(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2)$
តាង $\displaystyle x=t^a; y=t^b; z=t^c; t>0$
នោះយើងបានៈ
$7(t^{3a}+t^{3b}+t^{3c})+15t^{a+b+c}$$\geq 6t^{2a+b}+6t^{2b+c}+6t^{2c+a}+6t^{a+2b}+6t^{b+2c}+6t^{c+2a}$
$\displaystyle \Rightarrow \int\limits_{0}^{1}\frac{1}{t}(7t^{3a}+7t^{3b}+7t^{3c}+15t^{a+b+c})dt$$\displaystyle \geq \int\limits_{0}^{1}\frac{1}{t}(6t^{2a+b}+6t^{2b+c}+6t^{2c+a}+6t^{a+2b}+6t^{b+2c}+6t^{c+2a})dt$
$\displaystyle \Rightarrow \frac{7}{3a}+\frac{7}{3b}+\frac{7}{3c}+\frac{15}{a+b+c}$$\displaystyle \geq \frac{6}{2a+b}+\frac{6}{2b+c}+\frac{6}{2c+a}+\frac{6}{a+2b}+\frac{6}{b+2c}+\frac{6}{c+2a}$
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។ សមភាពកើតមានពេល $a=b=c$