Cho $a,b,c > 0$. Chứng minh rằng:
$6(a^{3}+b^{3}+c^{3})\geq 18abc+(\sum \sqrt[3]{a(b-c)^{2}})^{2}$
Edited by Baoriven, 25-09-2016 - 15:27.
CMR: $6(a^{3}+b^{3}+c^{3})\geq 18abc+(\sum \sqrt[3]{a(b-c)^{2}})^{2}$
Started By hoaichung01, 25-09-2016 - 12:57
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users
