Cho a,b,c là những số thực dương, CMR:
$\frac{b^{2}c^{3}}{a^{2}(b+c)^{3}}+\frac{c^{2}a^{3}}{b^{2}(c+a)^{3}}+\frac{a^{2}b^{3}}{c^{2}(a+b)^{3}}\geq \frac{9abc}{4(3abc+ab^2+bc^2+ca^2)}$
$\sum\frac{a^{2}b^{3}}{c^{2}(a+b)^{3}}\geq \frac{9abc}{4(3abc+ab^2+bc^2+ca^2)}$
Started By ngochapid, 16-01-2016 - 22:08
#2
Posted 21-07-2022 - 21:23
TARGET
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Binh nhì
Theo Holder bậc 4:
$\left ( \sum \frac{b^{2}c^{3}}{a^{2}\left ( b+c \right )^{3}} \right )\left ( 3abc+\sum ab^{2}\right )\left ( 2abc\sum a \right )\left ( 2\sum a \right )\geq \frac{\left ( \sum ab \right )^{4}}{\left ( 3abc+\sum ab^{2} \right )\left ( 2abc\sum a \right )\left ( 2\sum a \right )}$
Điều này tương đương$q^{4}\geq 9r^{2}p^{2}$
Luôn đúng theo BĐT AM-GM
Edited by TARGET, 21-07-2022 - 21:24.
$\sqrt[5]{\frac{a^{5}+b^{5}}{2}}\doteq \sqrt[5]{\frac{a^{5}+b^{5}}{a^{4}+b^{4}}\frac{a^{4}+b^{4}}{a^{3}+b^{3}}\frac{a^{3}+b^{3}}{a^{2}+b^{2}}\frac{a^{2}+b^{2}}{a+b}\frac{a+b}{2}}$
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