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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