efoski1687 31.07.2011 18:05 If $0\le a\le b\le c$ real numbers, prove that $(a+3b)(b+4c)(c+2a)\ge 60abc$.
123steeve 31.07.2011 18:41 AM-GM: \[LHS \ge 4a^{1/4}b^{3/4}\cdot5b^{1/5}c^{4/5}\cdot3c^{1/3}a^{2/3}=60[a^{55}b^{57}c^{68}]^{1/60}\ge 60abc\] since $c^{5} \ge a^{5}$ and $c^{3}\ge b^{3}$.
xeroxia 28.05.2012 18:33 There are other solutions at below links: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=147490109&t=104592 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=208321&p=1146593&hilit=60abc#p1146593
sqing 08.02.2015 00:47 See also here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=624353