# លំហាត់ 47: van khea

គេអោយ $a, b, c>0 ; abc=1$ ។ ស្រាយបញ្ជាក់ថាៈ
$\displaystyle \frac{1}{a^3(\sqrt[3]{b}+\sqrt[3]{c})}+\frac{1}{b^3(\sqrt[3]{c}+\sqrt[3]{a})}+\frac{1}{c^3(\sqrt[3]{a}+\sqrt[3]{b})}\geq \frac{3}{2}$
ចំលើយ
យើងមានៈ
$\displaystyle \sqrt[3]{a}+\sqrt[3]{b}\leq 2\sqrt[3]{\frac{a+b}{2}}=\sqrt[3]{4}\sqrt[3]{a+b}$
ដូចនេះយើងទាញបានៈ
$\displaystyle \frac{1}{a^3(\sqrt[3]{b}+\sqrt[3]{c})}+\frac{1}{b^3(\sqrt[3]{c}+\sqrt[3]{a})}+\frac{1}{c^3(\sqrt[3]{a}+\sqrt[3]{b})}$$\displaystyle \geq \frac{1}{\sqrt[3]{4}a^3\sqrt[3]{b+c}}+\frac{1}{\sqrt[3]{4}b^3\sqrt[3]{c+a}}+\frac{1}{\sqrt[3]{4}c^3\sqrt[3]{a+b}}$
ដូចនេះយើងត្រូវស្រាយថាៈ
$\displaystyle \frac{1}{\sqrt[3]{4}a^3\sqrt[3]{b+c}}+\frac{1}{\sqrt[3]{4}b^3\sqrt[3]{c+a}}+\frac{1}{\sqrt[3]{4}c^3\sqrt[3]{a+b}}\geq \frac{3}{2}$
$\displaystyle \Leftrightarrow \frac{1}{a^3\sqrt[3]{b+c}}+\frac{1}{b^3\sqrt[3]{c+a}}+\frac{1}{c^3\sqrt[3]{a+b}}\geq \frac{3}{\sqrt[3]{2}}$
យើងមានៈ
$\displaystyle \frac{1}{a^3\sqrt[3]{b+c}}+\frac{1}{b^3\sqrt[3]{c+a}}+\frac{1}{c^3\sqrt[3]{a+b}}$$\displaystyle =\frac{\biggl(\frac{1}{a}\biggl)^3}{1^{\frac{5}{3}}(b+c)^{\frac{1}{3}}}+\frac{\biggl(\frac{1}{b}\biggl)^3}{1^{\frac{5}{3}}(c+a)^{\frac{1}{3}}}+\frac{\biggl(\frac{1}{c}\biggl)^3}{1^{\frac{5}{3}}(a+b)^{\frac{1}{3}}}$
ដោយ $\displaystyle 3-\frac{5}{3}-\frac{1}{3}=1$ នោះតាមវិសមភាព $V - K:07$ គេបានៈ
$\displaystyle =\frac{\biggl(\frac{1}{a}\biggl)^3}{1^{\frac{5}{3}}(b+c)^{\frac{1}{3}}}+\frac{\biggl(\frac{1}{b}\biggl)^3}{1^{\frac{5}{3}}(c+a)^{\frac{1}{3}}}+\frac{\biggl(\frac{1}{c}\biggl)^3}{1^{\frac{5}{3}}(a+b)^{\frac{1}{3}}}$$\displaystyle \geq \frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^3}{3^{\frac{5}{3}}(2(a+b+c))^{\frac{1}{3}}}=\frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}{3^{\frac{5}{3}}(2(a+b+c))^{\frac{1}{3}}}$
យើងមានៈ $\displaystyle (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2\geq 3\biggl(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\biggl)=3(a+b+c)$
តាមវិសមភាព $AM - GM$ យើងមានៈ
$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{3}{\sqrt[3]{abc}}=3$
$\displaystyle \Rightarrow \frac{1}{a^3\sqrt[3]{b+c}}+\frac{1}{b^3\sqrt[3]{c+a}}+\frac{1}{c^3\sqrt[3]{a+b}}$$\displaystyle \geq \frac{9(a+b+c)}{3^{\frac{5}{3}}\sqrt[3]{2}(a+b+c)^{\frac{1}{3}}}=\frac{\sqrt[3]{3}(a+b+c)^{\frac{2}{3}}}{\sqrt[3]{2}}$
ដោយ $a+b+c\geq 3\sqrt[3]{abc}=3$
គេបានៈ
$\displaystyle \frac{1}{a^3\sqrt[3]{b+c}}+\frac{1}{b^3\sqrt[3]{c+a}}+\frac{1}{c^3\sqrt[3]{a+b}}\geq \frac{\sqrt[3]{3}.3^{\frac{2}{3}}}{\sqrt[3]{2}}=\frac{3}{\sqrt[3]{2}}$ ពិត ។
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់ ។ សមភាពកើតមានពលេ $a=b=c=1$

### 38 Responses to លំហាត់ 47: van khea

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