Toán 10 GTNN

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Cựu TMod Toán
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19 Tháng một 2019
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[tex]P=(\frac{1}{18}a+\frac{1}{24}b+\frac{2}{ab})+(\frac{1}{16}b+\frac{1}{8}c+\frac{2}{bc})+(\frac{2}{ac}+\frac{8}{abc}+\frac{1}{9}a+\frac{1}{9}a+\frac{1}{12}b+\frac{1}{6}c+\frac{1}{6}c)+\frac{13}{144}[(8a+6b)+(3b+6c)]\geq 3\sqrt[3]{\frac{1}{18}a.\frac{1}{24}b.\frac{2}{bc}}+3\sqrt[3]{\frac{1}{16}b.\frac{1}{8c}.\frac{2}{bc}}+7\sqrt[7]{\frac{2}{ac}.\frac{8}{abc}.\frac{1}{9}a.\frac{1}{9}a.\frac{1}{12}b.\frac{1}{6}c.\frac{1}{6}c}+\frac{13}{144}(2\sqrt{8a.6b}+2\sqrt{3b.6c})=\frac{1}{2}+\frac{3}{4}+\frac{7}{3}+\frac{13}{144}(2\sqrt{48ab}+2\sqrt{18bc})\geq \frac{43}{12}+\frac{13}{144}(2\sqrt{48.12}+2\sqrt{18.8})=\frac{43}{12}+\frac{13}{72}(24+12)=\frac{121}{12}[/tex]
 
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