Cho a,b>0. Chứng minh rằng $5\sqrt[5]{a}+12\sqrt[12]{b}\geq 17\sqrt[17]{ab}$
$5\sqrt[5]{a}+12\sqrt[12]{b}\geq 17\sqrt[17]{ab}$
Started By 12345678987654321123456789, 31-01-2016 - 00:29
#1
Posted 31-01-2016 - 00:29
12345678987654321123456789
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Even when you had two eyes, you'd see only half the picture.
#2
Posted 31-01-2016 - 10:52
tpdtthltvp
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Cho a,b>0. Chứng minh rằng $5\sqrt[5]{a}+12\sqrt[12]{b}\geq 17\sqrt[17]{ab}$
$5\sqrt[5]{a}+12\sqrt[12]{b}=\sqrt[5]{a}+\sqrt[5]{a}+\cdots +\sqrt[5]{a}+\sqrt[12]{b}+\sqrt[12]{b}+\cdots +\sqrt[12]{b}\geq 17\sqrt[17]{(\sqrt[5]{a})^5.(\sqrt[12]{b})^{12}}=17\sqrt[17]{ab}$
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$\color{red}{\mathrm{\text{How I wish I could recollect, of circle roud}}}$
$\color{red}{\mathrm{\text{The exact relation Archimede unwound ! }}}$
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