# 203 គណិតវិទ្យាសំរាប់សិស្សពូកែ

Van Khea: ស្រាយបញ្ជាក់ថាចំពោះបណ្ដាចំនួនវិជ្ជមាន $a, b, c$ គេបានៈ
$\displaystyle \frac{1}{a^3\sqrt{a^2+2b^2}}+\frac{1}{b^3\sqrt{b^2+2c^2}}+\frac{1}{c^3\sqrt{c^2+2a^2}}\geq \frac{\sqrt{3}}{abc\sqrt[3]{abc}}$
សំរាយបញ្ជាក់
យើងមានៈ
$\displaystyle \frac{1}{a^3\sqrt{a^2+2b^2}}+\frac{1}{b^3\sqrt{b^2+2c^2}}+\frac{1}{c^3\sqrt{c^2+2a^2}}$$\displaystyle =\frac{(\frac{1}{a^2})^{\frac{3}{2}}}{\sqrt{a^2+2b^2}}+\frac{(\frac{1}{b^2})^{\frac{3}{2}}}{\sqrt{b^2+2c^2}}+\frac{(\frac{1}{c^2})^{\frac{3}{2}}}{\sqrt{c^2+2a^2}}$
ដោយ $\displaystyle \frac{3}{2}-\frac{1}{2}=1$ នោះយើងបានៈ
$\displaystyle \frac{(\frac{1}{a^2})^{\frac{3}{2}}}{\sqrt{a^2+2b^2}}+\frac{(\frac{1}{b^2})^{\frac{3}{2}}}{\sqrt{b^2+2c^2}}+\frac{(\frac{1}{c^2})^{\frac{3}{2}}}{\sqrt{c^2+2a^2}}$$\displaystyle \geq \frac{(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})^{\frac{3}{2}}}{\sqrt{3(a^2+b^2+c^2)}}$
ម្យ៉ាងទៀតយើងមានៈ
$\displaystyle (\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})^{\frac{3}{2}}=((\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})^2)^{\frac{3}{4}}$$\displaystyle \geq (3(\frac{1}{a^2b^2}+\frac{1}{b^2c^2}+\frac{1}{c^2a^2}))^{\frac{3}{4}}$$\displaystyle =\frac{(3(a^2+b^2+c^2))^{\frac{3}{4}}}{(abc)^{\frac{3}{2}}}$
$\displaystyle \Rightarrow \frac{1}{a^3\sqrt{a^2+2b^2}}+\frac{1}{b^3\sqrt{b^2+2c^2}}+\frac{1}{c^3\sqrt{c^2+2a^2}}\geq \frac{(3(a^2+b^2+c^2))^{\frac{3}{4}}}{(abc)^{\frac{3}{2}}(3(a^2+b^2+c^2))^{\frac{1}{2}}}$$\displaystyle =\frac{(3(a^2+b^2+c^2))^{\frac{1}{4}}}{(abc)^{\frac{3}{2}}}$
តាមវិសមភាព $AM-GM$ យើងមានៈ $\displaystyle a^2+b^2+c^2\geq 3\sqrt[3]{a^2b^2c^2}$
$\displaystyle \Rightarrow \frac{1}{a^3\sqrt{a^2+2b^2}}+\frac{1}{b^3\sqrt{b^2+2c^2}}+\frac{1}{c^3\sqrt{c^2+2a^2}}\geq \frac{(3.3\sqrt[3]{a^2b^2c^2})^{\frac{1}{4}}}{(abc)^{\frac{3}{2}}}$$\displaystyle =\frac{\sqrt{3}}{abc\sqrt[3]{abc}}$
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។ សមភាពកើតមានពេល $a=b=c$