Cho $a,b,c>0$. CM
$P=\sum \frac{b^{2}c^{3}}{a^{2}+(b+c)^{3}}\geq \frac{9abc}{4(3abc+ab^{2}+bc^{2}+ca^{2})}$
Edited by tienduc, 10-02-2017 - 19:07.
CM $P=\sum \frac{b^{2}c^{3}}{a^{2}+(b+c)^{3}}\geq \frac{9abc}{4(3abc+ab^{2}+bc^{2}+ca^{2})}$
Started By tienduc, 10-02-2017 - 19:06
#2
Posted 10-02-2017 - 20:00
phamngochung9a
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Cho $a,b,c>0$. CM
$P=\sum \frac{b^{2}c^{3}}{a^{2}+(b+c)^{3}}\geq \frac{9abc}{4(3abc+ab^{2}+bc^{2}+ca^{2})}$
Bất đẳng thức sai với $a=b=c=1$ nhé bạn 
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